![Find the resistance of a copper coil of total wire-length 10 𝑚 and area
of cross-section 1.0 𝑚𝑚2
. What would be the resistance of a similar coil
of aluminium? The resistivity of copper = 1.7 × 10−8
Ω − 𝑚 and that of
aluminium = 2.6 × 10−8
Ω − 𝑚.
𝑅𝐶𝑢 =
𝜌𝐶𝑢 × 𝑙
𝐴
=
1.7 × 10−8
× 10
1 × 10−6
= 0.17 Ω
𝑅 =
𝜌𝑙
𝐴
⇒ 𝑅 ∝ 𝜌
𝑅𝐶𝑢
𝑅𝐴𝑙
=
𝜌𝐶𝑢
𝜌𝐴𝑙
⇒ 𝑅𝐴𝑙 =
𝜌𝐴𝑙
𝜌𝐶𝑢
× 𝑅𝐶𝑢 =
2.6 × 10−8
1.7 × 10−8
× 0.17 = 0.26 Ω
T
[if 𝑙 & 𝐴 are kept same]](https://image.slidesharecdn.com/currentelectricityjeeeng1-250804174639-cb1024a0/85/Current_Electricity_JEE_Eng-1-pdf-Ch-2-9-320.jpg)
![Current in the circuit,
𝑖 =
𝑁𝐸
𝑛1𝑟 + 𝑛2𝑅
=
𝑁𝐸
𝑛1𝑟 +
𝑁𝑅
𝑛1
Consider 𝑁 = 𝑛1𝑛2 identical cells, each of emf 𝐸 and internal resistance 𝑟. Suppose
𝑛1 cells are joined in series to form a line and 𝑛2 such lines are connected in
parallel. The combination drives a current in an external resistance 𝑅.
(b) Assuming that 𝑛1 and 𝑛2 can be continuously varied (𝑁 = 𝑛1𝑛2 still holds), find
the relation between 𝑛1, 𝑛2, 𝑅 and 𝑟 for which the current in 𝑅 is maximum.
When 𝑖 is maximum,
𝑖 =
𝑛1𝑁𝐸
𝑛1
2𝑟 + 𝑁𝑅
𝑑𝑖
𝑑𝑛1
= 0
𝑛1𝑟 = 𝑛2𝑅
⇒ 𝑛1
2
𝑟 + 𝑁𝑅 − 2𝑛1
2
𝑟 = 0
⇒ 𝑛1𝑛2𝑅 = 𝑛1
2
𝑟 (∵ 𝑁 = 𝑛1𝑛2)
⇒
⇒
𝑑𝑖
𝑑𝑛1
=
𝑁𝐸 𝑛1
2
𝑟 + 𝑁𝑅 − 𝑛1(𝑁𝐸)(2𝑛1𝑟)
[𝑛1
2𝑟 + 𝑁𝑅]2
= 0
∴ Condition for maximum current in 𝑅 is, 𝑛1𝑟 = 𝑛2𝑅](https://image.slidesharecdn.com/currentelectricityjeeeng1-250804174639-cb1024a0/85/Current_Electricity_JEE_Eng-1-pdf-Ch-2-54-320.jpg)
![Find the time constant for the given 𝑅𝐶 circuit.
[Given: 𝑅1 = 1 Ω, 𝑅2 = 2 Ω, 𝐶1 = 4 𝜇𝐹, 𝐶1 = 2 𝜇𝐹]
• Time constant for 𝑅𝐶 circuit :
• Equivalent resistance :
• Equivalent capacitance :
𝜏 = 𝑅𝑒𝑞𝐶𝑒𝑞
𝑅𝑒𝑞 = 𝑅1 + 𝑅2
𝑅𝑒𝑞 = 1 Ω + 2 Ω = 3 Ω
𝐶𝑒𝑞 = 𝐶1 + 𝐶2
𝐶𝑒𝑞 = 4 𝜇𝐹 + 2 𝜇𝐹 = 6 𝜇𝐹
(Resistors are in series)
(Capacitors are in parallel connection)
• Time constant:
𝜏 = 𝑅𝑒𝑞𝐶𝑒𝑞 = 3 × 6 = 18 𝜇𝑠 𝜏 = 18 𝜇𝑠
𝑖
1 Ω
2 Ω
4 𝜇𝐹
2 𝜇𝐹
E](https://image.slidesharecdn.com/currentelectricityjeeeng1-250804174639-cb1024a0/85/Current_Electricity_JEE_Eng-1-pdf-Ch-2-85-320.jpg)
Current_Electricity_JEE_Eng (1).pdf Ch 2
- 2. Electric Current Δ𝑡 Δ𝑄 𝒆 𝒆 𝒆 𝒆 • The net amount of charge flowing across the area in the time interval ∆𝑡, is defined to be the average current across the area. 𝑖𝑎𝑣𝑔 = Δ𝑄 Δ𝑡 • SI unit of electric current is 𝑎𝑚𝑝𝑒𝑟𝑒 (A). • Electric current has direction as well as magnitude but it is a scalar quantity. • If charge 𝑑𝑄 is flowing across the area in an infinitesimally small time interval 𝑑𝑡, it is defined to be the instantaneous current across that area. 𝑖𝑖𝑛𝑠𝑡 = 𝑑𝑄 𝑑𝑡
- 3. Direction of Conventional Electric Current • In reality, the free electrons of the conductor flow from end B at lower potential to end A at higher potential. • However, we choose the direction of electric current as that of supposed flow of positive charge as convention. i.e., from A to B. 𝒆 𝒆 𝒆 𝒆 𝐵 𝐴 𝐼 𝐹𝑙𝑜𝑤 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝒆 𝒆 𝑉𝐴(> 𝑉𝐵) 𝑉𝐵 + -
- 4. T T Given 𝑖 = (2𝑡 + 3𝑡2 ) 𝐴 To find 𝑄 during 𝑡 = 2 𝑠 to 𝑡 = 3 𝑠 Charge flowing in a time interval is given by: 𝑄 = න 2 3 𝑖 𝑑𝑡 ⇒ 𝑄 = න 2 3 (2𝑡 + 3𝑡2 ) 𝑑𝑡 ∴ 𝑄 = 24 𝐶 ⇒ 𝑄 = 𝑡2 2 3 + 𝑡3 2 3 The current in a conductor varies with time 𝑡 as, 𝑖 = 2𝑡 + 3𝑡2 𝐴. where 𝑖 is in ampere and 𝑡 is in second. Find the electric charge flowing through a section of the conductor during 𝑡 = 2 𝑠 to 𝑡 = 3 𝑠.
- 5. Electric Current Density ∆𝑆 𝑖 𝑗𝑎𝑣𝑔 = Δ𝑖 Δ𝑆 𝑗𝑖𝑛𝑠𝑡 = 𝑑𝑖 𝑑𝑆 • A vector quantity whose magnitude is equal to electric current per unit normal cross-sectional area inside a conductor. • Under electrostatic conditions, Ԧ 𝑗 = 0 • If the current ∆𝑖 exists through an area ∆𝑆 which makes an angle 𝜃 with the current, as shown in the figure, then The current density: 𝑗 = ∆𝑖 ∆𝑆 cos 𝜃 ∆𝑖 = Ԧ 𝑗. ∆ Ԧ 𝑆 • For non-uniform current density, 𝑖 = ∫ Ԧ 𝑗. 𝑑 Ԧ 𝑆 , Ԧ 𝑗 → Uniform current density
- 6. Drift Speed • After application of an electric field, average velocity attained by electrons in a material is known as drift speed. 𝑣𝑎𝑣𝑔 = 𝑣𝑑 = 𝑒𝐸 𝑚 𝜏 • The average time interval between two successive electron collisions is called relaxation time(𝜏). • In a conductor, the free electrons exhibit a random motion (known as Brownian motion). During this motion, the electrons collide with the heavy fixed ions. But after collision, electrons emerge with same speed but in random directions.
- 7. 𝑣𝑑Δ𝑡 𝐸 𝑣𝑑 𝐴 Electron 𝑛 = No. of free electrons per unit volume or Mobile electron density 𝑖 = 𝑛𝑒𝐴𝑣𝑑 𝑗 = 𝑛𝑒𝑣𝑑 • Electric current, • Current density, • Physically, the mobility (𝜇) refers to the ease with which the charge carriers move in the medium. 𝜇 = 𝑣𝑑 𝐸 = 𝑒𝜏 𝑚 Mobility Ԧ 𝑗 = 𝜎𝐸 By substituting the value of 𝑣𝑑 in above equation, we get, 𝑗 = 𝑛𝑒𝑣𝑑 = 𝑛𝑒2 𝜏 𝑚 𝐸 = 𝜎𝐸 , 𝜎 = electrical conductivity of material 𝜌 = 1 𝜎 = 𝑚 𝑛𝑒𝜏2 Resistivity,
- 8. Ohm’s Law 𝑙 𝐴 𝐼 𝐸 𝑉 • Resistance of a given cylindrical conductor, • The current through a conductor(𝐼) is directly proportional to the potential difference(𝑉) applied across its ends. 𝑅 = 𝜌𝑙 𝐴 𝑉 = 𝑅𝐼 𝑅 = 𝜌𝑏 𝑎𝑐 𝑅 = 𝜌𝑐 𝑏𝑎 𝑅 = 𝜌𝑎 𝑏𝑐 Resistance of a conductor depends on: • Resistivity of material • Dimensions
- 9. Find the resistance of a copper coil of total wire-length 10 𝑚 and area of cross-section 1.0 𝑚𝑚2 . What would be the resistance of a similar coil of aluminium? The resistivity of copper = 1.7 × 10−8 Ω − 𝑚 and that of aluminium = 2.6 × 10−8 Ω − 𝑚. 𝑅𝐶𝑢 = 𝜌𝐶𝑢 × 𝑙 𝐴 = 1.7 × 10−8 × 10 1 × 10−6 = 0.17 Ω 𝑅 = 𝜌𝑙 𝐴 ⇒ 𝑅 ∝ 𝜌 𝑅𝐶𝑢 𝑅𝐴𝑙 = 𝜌𝐶𝑢 𝜌𝐴𝑙 ⇒ 𝑅𝐴𝑙 = 𝜌𝐴𝑙 𝜌𝐶𝑢 × 𝑅𝐶𝑢 = 2.6 × 10−8 1.7 × 10−8 × 0.17 = 0.26 Ω T [if 𝑙 & 𝐴 are kept same]
- 10. Ohmic & Non-ohmic Materials Validity of Ohm’s law Ohmic Materials Non-ohmic Materials 𝐼 𝑉 Slope = 1/𝑅 𝐼 𝑉 • Metals at low 𝑉 and 𝐼 (Temperature and other physical conditions must remain constant) Ex. Nichrome • Semiconductors & alloys.
- 11. A conductor of length 𝑙 has a circular cross section as shown. The radius of cross section varies from 𝑎 to 𝑏. The resistivity of the material is 𝜌. Assume that 𝑏 − 𝑎 ≪ 𝑙, find the resistance of the conductor across the ends 𝑃 and 𝑄. ⇒ 𝑑𝑅 = 𝜌𝑑x 𝜋𝑟2 Radius of element at 𝑥 distance from end 𝑃, ⇒ 𝑟 = 𝑎 + 𝑏 − 𝑎 𝑙 𝑥 Resistance of the conductor, ⇒ 𝑅 = න 𝑎 𝑏 𝜌 𝜋𝑟2 𝑙 𝑏 − 𝑎 𝑑𝑟 ⇒ 𝑅 = 𝜌𝑙 𝜋𝑎𝑏 𝑙 𝑎 𝑏 𝜃 𝑥 𝑑𝑥 𝑟 𝑄 𝑃 We know resistance of a cylindrical conductor, 𝑅 = 𝜌𝑙 𝐴 For elemental disc, Solution:
- 12. The masses of the wires of copper are in the ratio 1: 3: 5 and their lengths are in the ratio 5: 3: 1. The ratio of their electrical resistance is ⇒ 𝑅 = 𝜌𝑙2 𝑑 𝑀 𝑀 = 𝐴𝑙 × 𝑑 ⇒ 𝑅 ∝ 𝑙2 𝑀 Ratio of 𝑅1, 𝑅2 & 𝑅3, ∴ 𝑅1 ∶ 𝑅2 ∶ 𝑅3 = 52 1 : 32 3 : 12 5 𝑅1 ∶ 𝑅2 ∶ 𝑅3 = 125 ∶ 15 ∶ 1 Solution: 𝑀1: 𝑀2: 𝑀3 = 1: 3: 5, 𝑙1: 𝑙2: 𝑙3 = 5: 3: 1 𝐴3 𝑙3, 𝑀3 𝐴1 𝑙2, 𝑀2 𝐴2 𝑙1, 𝑀1 We know resistance of a cylindrical conductor, 𝑅 = 𝜌𝑙 𝐴
- 13. Resistivity Dependence Factors – Types of Material Conductor Semiconductor Insulator 𝜌 ≈ 10−12 − 10−6 Ω − 𝑚 𝜌 ≈ 10−5 − 105 Ω − 𝑚 𝜌 ≈ 105 − 1016 Ω − 𝑚 𝜌 = 𝑚 𝑛𝑒2𝜏
- 14. Resistivity Dependence Factors – Temperature Change in resistivity with temperature : 𝜌 = 𝜌0 1 + 𝛼 𝑇 − 𝑇0 𝜌 = resistivity at temperature 𝑇 𝜌0 = resistivity at temperature 𝑇0 𝛼 = temperature coefficient of resistivity Resistance of a conductor at temperature 𝑇 : 𝑅 = 𝑅0 1 + 𝛼 𝑇 − 𝑇0 Effect of Temperature on Resistivity of Different Materials Temperature (𝑇) Resistivity (𝜌), For Conductors 𝛼 = +𝑣𝑒 Ω − 𝑚 For Semiconductors Temperature (𝑇) Resistivity 𝜌 , 𝛼 = −𝑣𝑒 Ω − 𝑚 °𝐶 °𝐶
- 15. T The temperature coefficient of resistance of wire is 0.00125 °𝐶−1 . At 300 𝐾 its resistance is 1 Ω. At what temperature will its resistance become 2Ω ? Let the resistance of wire is 𝑅0 at 𝑇 = 0 °𝐶. Hence, 𝑅1 = 𝑅0 1 + 𝛼𝑇1 & 𝑅2 = 𝑅0(1 + 𝛼𝑇2) ⇒ 𝑅1 𝑅2 = 1 + 𝛼𝑇1 1 + 𝛼𝑇2 ⇒ 𝛼 = 𝑅1 − 𝑅2 𝑅2𝑇1 − 𝑅1𝑇2 1.25 × 10−3 = 1 − 2 2 27 − 1(𝑇2) ⇒ 𝑇2 = 854 °𝐶 𝑇2 = 1127 𝐾 or Solution : 𝑅1 = 1 𝛺, 𝑅2 = 2 𝛺, 𝑇1 = 300 𝐾 = 27 °𝐶, 𝛼 = 0.00125 °𝐶−1
- 16. Resistor and Colour Coding • Resistor is an object with desired resistance. Symbol of resistor Resistor 1 2 3 4 Indicate the first and second significant figures of the resistance in ohm. Colour 4 or no colour stands for tolerance or possible variation in percentage about the indicated values. Indicates the decimal multiplier.
- 17. Colour Digit Multiplier Tolerance Black 0 1 Brown 1 101 Red 2 102 Orange 3 103 Yellow 4 104 Green 5 105 Blue 6 106 Violet 7 107 Grey 8 108 White 9 109 Gold 10−1 ± 5% Silver 10−2 ± 10% No colour ± 20% Colour Coding of Carbon Resistors Trick B B R O Y Great Britain has a Very Good Wife Example: Resistance of Carbon resistor, 𝑅 = 53 × 104 ± 5 % Digit 1 = 5 (Green) Digit 2 = 3 (Orange) Multiplier = 104 (Yellow) Tolerance = ± 5 % (Gold) 𝑅 = 530 𝑘Ω ± 5 %
- 18. Ohm’s law & Sign Convention • Sign Convention : Case 1: 𝑉𝐴 > 𝑉𝐵 Case 2: 𝑉𝐴 < 𝑉𝐵 𝐼 − + 𝑉𝐴 𝑉𝐵 𝐼 + − 𝑉𝐴 𝑉𝐵 ⇒ 𝑉𝐴 − 𝑉𝐵 = 𝐼𝑅 ⇒ 𝑉𝐵 − 𝑉𝐴 = 𝐼𝑅 • Current, by convention is always from higher to lower potential. + 𝑣𝑒 sign is taken for higher potential and − 𝑣𝑒 sign for lower potential. Along path 𝐴 → 𝐵 Along path 𝐴 → 𝐵 𝑉𝐴 − 𝐼𝑅 = 𝑉𝐵 𝑉𝐴 + 𝐼𝑅 = 𝑉𝐵 • The current through a conductor (𝐼) is directly proportional to the potential difference(𝑉) applied across its ends. 𝑉 = 𝑅𝐼
- 19. • Electric cell is a device which maintains potential difference between its terminals. • Converts chemical energy into electrical energy. Symbolic representation of a cell Electric Cell (Battery) • EMF of a battery is defined as the work done (𝑊) in driving a unit charge (𝑞) across the terminals of the cell. • EMF is also written as the potential difference across the cell when it is not connected to the external circuit. • 𝑟 is internal resistance of the cell. • Potential difference across an ideal cell is equal to its EMF. + − 𝑖 𝑟 𝐴 𝐵 Electrolyte EMF of Electric Cell
- 20. Discharging(Source) Charging (Load) 𝑖 𝐴 𝐵 𝑟 E + − 𝑉𝐴𝐵 = E − 𝑖𝑟 𝑉𝐴𝐵 = E + 𝑖𝑟 • Terminal Voltage is less than EMF of cell. • Terminal Voltage is Greater than EMF of cell. Terminal Voltage of a Cell 𝑖 𝐴 𝐵 𝑟 E + −
- 21. In the diagram shown below , find: 𝑎 𝑉𝐴𝐵 𝑏 𝑉𝐶𝐴 in steady state. 2 Ω 4 Ω 𝐴 8 𝑉 10 𝑉 𝐶 𝐵 2 𝐴 • All elements are in a line. Hence current through all of them is 2 𝐴. • Potential drop across each resistor, 𝑉2Ω = 2 × 2 = 4 𝑉 & 𝑉4Ω = 2 × 4 = 8 𝑉 Potential at every point. 𝑉𝐴𝐵 = −10 𝑉 𝑉𝐶𝐴 = 16 𝑉 2 Ω 4 Ω 𝐴 8 𝑉 10 𝑉 𝐶 𝐵 2 𝐴 2 𝐴 𝑉𝐵 = 0 𝐷 𝐸 𝑉𝐷 = 10 𝑉𝐶 = 6 𝑉 𝑉𝐸 = −2 𝑉 𝑉𝐴 = −10 𝑉 + − + − 8 𝑉 4 𝑉
- 22. Kirchhoff’s Laws Kirchhoff’s Current Law(KCL) Kirchhoff’s Voltage Law(KVL) 𝐴 𝐵 𝐶 𝐷 ℰ 𝑅1 𝑅2 𝑅3 + − + − + − + − 𝐼 • Incoming current = Outgoing current • Net current at a junction = 0 𝑖1 + 𝑖2 − 𝑖3 = 0 • At the junction ⇒ 𝑄1 + 𝑄2 = 𝑄3 Sum of charges entering the node Sum of charges leaving the node = • KCL is based on the conservation of charge principle . 𝐴 𝐵 𝐶 𝑅1 𝑅2 𝑅3 𝑖1 𝑖2 𝑖3 • In a closed loop, the algebraic sum of all the potential differences is zero. • KVL is Based on energy conservation . • Apply KVL from Point 𝐶, +E − 𝑖𝑅1 − 𝑖𝑅2 − 𝑖𝑅3 = 0 ⇒ E = 𝑖 (𝑅1 + 𝑅2 + 𝑅3) 𝑖1 + 𝑖2 = 𝑖3
- 23. Sign convention for KVL For Battery/Cell For Resistor Δ𝑉 → −𝑣𝑒 • If we choose to traverse along the direction of current , then 𝐼 − + • If we choose to traverse opposite to the direction of current then, Δ𝑉 → +𝑣𝑒 𝐼 − + • If you encounter −𝑣𝑒 terminal of the cell first, then, Δ𝑉 → +𝑣𝑒 E + − • If you encounter +𝑣𝑒 terminal of the cell first, then, Δ𝑉 → −𝑣𝑒 E + −
- 24. In the electrical circuit shown below, find the value of 𝑖. 𝐴 𝐵 𝐶 1 𝐴 2 𝐴 4 𝐴 𝑖 𝐷 Incoming current Outgoing current = ⇒ 4 + 2 = 1 + 𝑖 5 𝐴 ⇒ 𝑖 = Assuming direction of 𝑖 as shown & using KCL ,
- 25. The potential difference (𝑉𝐴 − 𝑉𝐵) across the points 𝐴 and 𝐵 in the given figure is _____. +3 𝑉 A −3 𝑉 B +9 𝑉 C −9 𝑉 D 2 𝐴 2 Ω 3 𝑉 1 Ω 𝐵 𝐴 𝑉𝐴 𝑉𝐵 Applying KVL from 𝐴 to 𝐵 𝑉𝐴 − 2 2 − 3 − 2 1 = 𝑉𝐵 𝑉𝐴 − 𝑉𝐵 = + 9 𝑉
- 26. In the circuit shown below, the conductor 𝑋𝑌 is of negligible resistance. Then find the current through 𝑋𝑌. 𝑅1 𝑋 𝑅2 𝑌 E1 E2 • In a loop entry and exit current of a cell always remains same. • If some amount of current flows through 𝑋𝑌, then enter and exit current will not be same for both the cells. Therefore, KCL will be violating. No Current will flow through 𝑋𝑌
- 27. Each of the resistors shown in figure has a resistance of 10 Ω and each of the batteries has an emf of 10 𝑉. Find the currents through the resistors 𝑅1 and 𝑅2 in following circuits: 10 𝑉 𝐴 𝑅1 𝑅2 10 𝑉 𝐵 𝐶 𝑖1 𝑖2 • Step 1 : Choose zero potential at any point. • Step 2 : Write potential of every point 10 𝑉 0 𝑅1 𝑅2 10 𝑉 10 𝑉 𝐶 𝑖1 𝑖2 0 0 0 0 10 𝑉 𝐵 𝐴 • Current through resistor 𝑅1 & 𝑅2. 𝑖1 = 1 𝐴 & 𝑖2 = 0 𝐴 10 𝑉 𝑅1 𝑅2 10 𝑉 𝐶 𝑖1 𝑖2 0 𝐵 𝐴
- 28. Find the potential difference 𝑉𝐴 − 𝑉𝐵. In the Following circuit. E1 𝑅1 E2 𝑅3 𝑅2 𝐴 𝐵 E1 𝑖1 𝑅1 E2 𝑅3 𝑅2 𝐴 𝐵 𝑖2 𝑖1 + 𝑖2 0 0 0 𝑉 𝑉 𝑉 V−E1 V−E2 • Since we have to find 𝑉𝐴 − 𝑉𝐵, always try to take 𝑉𝐵 = 0 for this type of the question, so that 𝑉 will be equal to 𝑉𝐴 • Apply KCL at point 𝐵. 𝑉 − E1 − 0 𝑅1 + 𝑉 − 0 𝑅3 + 𝑉 − E2 − 0 𝑅2 = 0 ⇒ 𝑉 = 𝑉𝐴 − 𝑉𝐵 = E1 𝑅1 + E2 𝑅2 1 𝑅1 + 1 𝑅2 + 1 𝑅3
- 29. Series Combination Parallel Combination Combination of Resistances • Current is same through all resistors. • For 𝑛 number of resistances in series, 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 𝑅𝑒𝑞 = 𝑅𝑖 𝐴 𝐵 𝑅𝑒𝑞 𝐶 𝐷 E 𝑖 𝐴 𝐵 𝑅2 𝐶 𝐷 𝑅1 𝑅3 E 𝑖 𝐴 𝐵 𝑅𝑒𝑞 𝑖 𝑖 𝑖 E 𝑖 𝐴 𝑅2 𝐵 𝑅1 𝑅3 E 𝑖3 𝑖2 𝑖1 • Potential difference is same across all resistors. • For 𝑛 number of resistances in parallel, 1 𝑅𝑒𝑞 = 1 𝑅1 + 1 𝑅2 + 1 𝑅3 + ⋯
- 30. Find the effective resistance between the points 𝐴 and 𝐵 of the network as shown in figure. 3 Ω 6 Ω 6 Ω 6 Ω 3 Ω 3 Ω 3 Ω 𝐴 3 Ω 3 Ω 𝐵 𝐷 𝐸 𝐹 𝐶 𝟑 𝛀 𝟑 𝛀 𝐹 Series 𝟔 𝛀 𝟔 𝛀 𝐴 𝐸 Parallel 𝟑𝛀 3 𝛀 𝐴 Series 3 Ω 𝐴 𝐵 6 Ω RAB = 2 Ω
- 31. In the circuit shown below, find the equivalent resistance across 𝐴 and 𝐵. 6 Ω 6 Ω 12 Ω 4 Ω 3 Ω 𝐴 𝐵 𝐴 𝐴 𝐴 𝐵 𝐵 𝐵 𝐴 𝐴 𝐵 1 𝑅𝑒𝑞 = 1 3 + 1 6 + 1 12 + 1 4 + 1 6 = 4 + 2 + 1 + 3 + 2 12 ⇒ 𝑅𝑒𝑞 = 1 Ω 4 Ω 12 Ω 6 Ω 𝐴 𝐵 3 Ω 6 Ω
- 32. Find the equivalent resistance of the network between the points a and b when; (a) The switch 𝑆 is open and (b) The switch 𝑆 is closed. 6 Ω 12 Ω 6 Ω 12 Ω 𝑎 𝑏 𝑎 𝑏 𝑅𝑎𝑏 = 9 Ω 𝑅1 = 18 Ω 𝑎 𝑏 𝑅2 = 18 Ω ⇒ 𝑅𝑎𝑏 = 9 Ω 𝑅1 = 6 + 12 = 18 Ω 𝑅2 = 12 + 6 = 18 Ω 1 𝑅𝑎𝑏 = 1 𝑅1 + 1 𝑅2 Step - 1 Step - 2 ⇒ 1 𝑅𝑎𝑏 = 2 18 6 Ω 12 Ω 6 Ω 12 Ω 𝑎 𝑏 𝑆 𝑎 𝑏 𝑅𝑎𝑏 = 8 Ω 𝑅1 = 4 Ω 𝑎 𝑏 𝑅2 = 4 Ω 𝑅1 = 6 × 12 6 + 12 = 4 Ω 𝑅𝑎𝑏 = 8 Ω 𝑅𝑎𝑏 = 𝑅1 + 𝑅2 𝑅2 = 6 × 12 6 + 12 = 4 Ω Step - 1 Step - 2 ⇒ 𝑅𝑎𝑏 = 4 + 4
- 33. 𝐴 and 𝐵 are two points on a uniform ring of resistance 𝑅. The ∠𝐴𝐶𝐵 = 𝜃, where 𝐶 is the centre of the ring. The equivalent resistance between 𝐴 and 𝐵 is 𝐵 𝜃 𝐴 𝐶 ≡ • Major & Minor arc are trapped between same points hence they exists in parallel combination. 𝑅𝑒𝑞 = 𝑅𝜃(2𝜋 − 𝜃) 4𝜋2 𝐴 𝐵 1 𝑅𝑒𝑞 = 1 𝑅1 + 1 𝑅2 = 𝑅1 × 𝑅2 𝑅1 + 𝑅2 𝑅1 = 𝑅 2𝜋 𝜃 𝑅2 = 𝑅 2𝜋 (2𝜋 − 𝜃)
- 34. Find the currents in the different resistors shown in figure. 2 𝑉 8 Ω 8 Ω 2 Ω 2 Ω 4 Ω 2 𝑉 2 𝑉 Applying KCL at junction 𝐴, we get, 𝑉 − 0 4 + 𝑉 − 0 4 + 𝑉 − 0 8 + 𝑉 − 0 8 = 0 𝑉 = 0 Current through all resistances is zero
- 35. A conducting wire of resistance 𝑅 is bent to form a square 𝐴𝐵𝐶𝐷 as shown in the figure. Find the effective resistance between 𝐸 and 𝐶 (𝐸 is mid-point of arm 𝐶𝐷). 𝐴 𝐵 𝐶 𝐷 𝐸 𝑅1 𝑅2 𝐸 𝐶 𝑅1 𝑅2 • Length of each side of the square = 𝑎 • Resistance per unit length of the square = 𝑅 4𝑎 • Resistance, 𝑅1 = 𝑅 4𝑎 × 𝐸𝐶 = 𝑅 4𝑎 × 𝑎 2 = 𝑅 8 • Resistance, 𝑅2 = 𝑅 4𝑎 × (𝐸𝐷𝐴𝐵𝐶) = 𝑅 4𝑎 × 𝑎 2 + 3𝑎 = 7𝑅 8 • Effective resistance between 𝐸 and 𝐶 : 𝑅𝑒𝑞 = 𝑅1 × 𝑅2 𝑅1 + 𝑅2 = 𝑅 8 × 7𝑅 8 𝑅 8 + 7𝑅 8 = 7𝑅 64
- 36. Wheatstone Bridge 𝑅1𝑅4 = 𝑅2𝑅3 Or 𝑅2 𝑅1 = 𝑅4 𝑅3 𝑖𝐺 = 0 Balanced Unbalanced The wheatstone bridge will be unbalanced if, • Equivalent resistance : 𝑅2 𝑅1 ≠ 𝑅4 𝑅3 𝑅𝑒𝑞 = E 𝑖 • Equivalent resistance : 𝑅𝑒𝑞 = 𝑅2 + 𝑅4 𝑅1 + 𝑅3 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 The wheatstone bridge will be balanced if,
- 37. Find the equivalent resistance of the shown network between points 𝑎 and 𝑏. 𝑎 𝑏 𝑅 𝑅 𝑅 𝑅 𝑅 𝑎 𝑏 𝑅 𝑅 𝑅 𝑅 𝑅 • The network is a balanced wheatstone bridge. 𝑅𝑒𝑞 = 2𝑅 × 2𝑅 2𝑅 + 2𝑅 = 𝑅 𝑎 𝑏 𝑅 𝑅 𝑅 𝑅
- 38. Find equivalent resistance between points 𝑎 and 𝑏. • Assuming an external battery of E supplies 𝑖 current in the circuit : 𝑅𝑒𝑞 = E 𝑖 1 • Apply KVL for loop 1 : −10𝑖1 − 5 2𝑖1 − 𝑖 + 5 𝑖 − 𝑖1 = 0 • Apply KVL for external loop 𝑎𝑐𝑑𝑏𝑒𝑓𝑎 : −10𝑖1 − 5 𝑖 − 𝑖1 + E = 0 => E = 5 2𝑖 5 + 𝑖 𝑖1 = 2𝑖 5 => E = 5 𝑖1 + 𝑖 𝑅𝑒𝑞 = E 𝑖 = 7 Ω The wheatstone bridge is unbalanced. • The current through 𝑅1 and 𝑅4 will be same, and the current through 𝑅2 and 𝑅3 will be same. • It possesses input/output symmetry. 10 5 ≠ 5 10
- 39. In a Wheatstone bridge a battery of 2𝑉 is used as shown in the figure. Find the current through middle branch. • The wheatstone bridge is unbalanced since 1 2 ≠ 2 3 • Apply KCL at node 𝐵 : 𝑥 − 0 2 + 𝑥 − 𝑦 4 + 𝑥 − 2 1 = 0 • Apply KCL at node 𝐷 : ⇒ 13𝑦 − 3𝑥 = 12 ⇒ 7𝑥 − 𝑦 = 8 𝐴 𝐶 1 Ω 2 Ω 3 Ω 2 Ω 4 Ω 𝐵 𝐷 2 𝑉 2 0 𝑥 𝑦 𝑦 − 0 3 + 𝑦 − 𝑥 4 + 𝑦 − 2 2 = 0 …….(1) …….(2) 𝑦 = 27 22 𝑉 𝑥 = 203 154 𝑉 Current through middle branch : 𝐼4Ω = 𝑥 − 𝑦 4 𝐼4Ω = 7 308 𝐴 Solving (1) and (2) we get:
- 40. • Null point : A point on the wire of the meter bridge for which no deflection will be shown by the needle of the galvanometer. • It is an electrical instrument used to find the value of unknown resistance. It consists of a 1 𝑚 (100 𝑐𝑚) wire with uniform cross-section. Meter Bridge • It works based on the principle of “Balanced Wheatstone Bridge”. • Condition of null-point : • Unknown resistance : 𝑅 = 𝑆 𝑙 100 − 𝑙 𝑅 𝑆 = 𝑙 100 − 𝑙
- 41. A meter bridge is set-up, as shown, to determine an unknown resistance ‘𝑋’ using a standard 10 Ω resistor. The galvanometer shows null point when tapping-key is at 52 𝑐𝑚 mark. The determined value of ‘𝑋’ is Distance of null point 𝑃 from end 𝐴 is, 𝐴𝑃 = 52 𝑐𝑚 Distance of null point 𝑃 from end 𝐵 is, 𝑃𝐵 = 100 − 52 = 48 𝑐𝑚 Unknown resistance: 𝑋 = 10 × 52 48 𝑋 = 10.8 Ω
- 42. Find equivalent resistance between points 𝑎 and 𝑏 of the infinite ladder network of resistances. • Equivalent resistance between 𝑎 and 𝑏 is, 𝑅𝑎𝑏 ≡ 𝑅𝑒𝑞 = 1 + 2 × 𝑅𝑒𝑞 2 + 𝑅𝑒𝑞 𝑅𝑒𝑞 = 3𝑅𝑒𝑞 + 2 2 + 𝑅𝑒𝑞 𝑅𝑒𝑞 2 − 𝑅𝑒𝑞 − 2 = 0 𝑅𝑒𝑞 = 2 Ω
- 43. Symmetric Circuit 𝑉 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑖 𝑖 − 2𝑥 𝑖 − 2𝑥 𝑃 𝑄 𝑋 𝑌 𝑂 𝑥 𝑥 𝑥 𝑥 𝑉 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑖 𝑖 − 2𝑥 𝑖 − 2𝑥 𝑃 𝑄 𝑋 𝑌 𝑥 𝑥 𝑥 𝑥 • The circuit has input & output symmetry & junction 𝑂 is redundant. 𝑉 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑖 𝑖1 𝑖1 𝑃 𝑄 𝑋 𝑌 𝑂 𝑖2 𝑖2 𝑖3 𝑖3 Find equivalent resistance of the circuit shown.
- 44. Symmetric Circuit 𝑉 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑖 𝑖 − 2𝑥 𝑖 − 2𝑥 𝑃 𝑄 𝑋 𝑌 𝑥 𝑥 𝑥 𝑥 𝑉 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 2𝑅 3 2𝑅 3 𝑖 𝑃 𝑄 𝑋 𝑌 𝑅𝐴𝑃𝑄𝐵 = 𝑅 + 2𝑅 3 + 𝑅 = 8𝑅 3 𝑅𝐴𝑋𝑌𝐵 = 𝑅 + 2𝑅 3 + 𝑅 = 8𝑅 3 𝑅𝐴𝐵 = 2𝑅 1 𝑅𝑒𝑞 = 3 8𝑅 + 3 8𝑅 + 1 2𝑅 = 10 8𝑅 = 5 4𝑅 𝑅𝑒𝑞 = 4𝑅 5
- 45. Find the equivalent resistance of following circuit between 𝐴 and 𝐵 , given that each resistance is 𝑅. 𝐴 𝐵 𝑃 𝑄 𝑖/6 𝑖/6 E 𝑖/3 𝑖/3 𝑖/3 𝑖/3 𝑖/3 𝑖/3 𝑖 Applying KVL in the loop 𝐴𝑃𝑄𝐵𝐴 on starting from point 𝑃, we get, − 𝑖𝑅 6 − 𝑖𝑅 3 + E− 𝑖𝑅 3 = 0 ⇒ E 𝑖 = 5𝑅 6 ⇒ 𝑅𝑒𝑞 = 5𝑅 6
- 46. Find the equivalent resistance of following circuit between 𝐴 and 𝐵 , given that each resistance is 𝑅. 𝐴 𝐵 𝐴1 𝐵1 𝑃 𝑃1 𝑄1 𝑄 E 𝑥 𝑖 − 2𝑥 𝑥 𝑥 𝑥 𝑖 − 2𝑥 𝐴 𝐵 𝐴1 𝐵1 𝑃 𝑃1 𝑄1 𝑄 E • The circuit has input & output symmetry & junction 𝑃 & 𝑄 are redundant. Therefore current through branches 𝑃𝑃1 & 𝑄𝑄1 are zero. 1 𝑅𝑒𝑞 = 1 3𝑅 + 1 𝑅 ⇒ 𝑅𝑒𝑞 = 3𝑅 4 𝐴 𝐵 E 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝑅 𝐴 𝐵 𝑅 𝑅 𝑅 𝑅
- 47. Find the equivalent resistance of following circuit between 𝐴 and 𝐵 , given that each resistance is 𝑅. 𝐴 𝐵 E 𝐴 𝐵 E 𝑃 𝑄 𝑅 𝑆 𝑇 𝑈 𝑥 𝑥 𝑖 − 2𝑥 𝑖 𝑥 𝑥 𝑦 𝑥 − 𝑦 2(𝑥 − 𝑦) 𝑥 − 𝑦 𝑥 − 𝑦 𝑦 𝑥 − 𝑦
- 48. Applying KVL in loop 𝑃𝑆𝑇𝑅𝑃, − 𝑥 − 𝑦 𝑅 − 2 𝑥 − 𝑦 𝑅 − 𝑥 − 𝑦 𝑅 + 𝑦𝑅 = 0 ⇒ 4𝑥 = 5𝑦 … … (1) Applying KVL in loop 𝐴𝑄𝑈𝐵𝐴, −𝑥𝑅 − 𝑦𝑅 − 𝑥𝑅 + 𝑖 − 2𝑥 𝑅 = 0 ⇒ 4𝑥 + 𝑦 = 𝑖 … … (2) Applying KVL in loop 𝐴𝐵𝐴, 𝑖 − 2𝑥 𝑅 = E … … (3) From equation 1 & (2), 4𝑥 + 4𝑥 5 = 𝑖 ⇒ 𝑥 = 5𝑖 24 Substituting value of 𝑥 in equation (3), 𝑖 − 5𝑖 12 𝑅 = E ⇒ 𝑅𝑒𝑞 = E 𝑖 = 7𝑅 12
- 49. Series Combination Parallel Combination Combination of Cells E𝑒𝑞 = E𝑖 𝑟𝑒𝑞 = σ 𝑟𝑖 Algebraic addition with sign Net resistance, Net emf = E𝑒𝑞 = E2𝑟1 + E1𝑟2 𝑟1 + 𝑟2 E𝑒𝑞 = 𝑟𝑒𝑞 E1 𝑟1 + E2 𝑟2 1 𝑟𝑒𝑞 = 1 𝑟1 + 1 𝑟2 Net emf = Net resistance,
- 50. Series Combination Parallel Combination Combination of Cells Total resistance = 𝑛𝑟 + 𝑅 Net emf = 𝑛 E Current in circuit, 𝑖 = 𝑛 𝑛𝑟 + 𝑅 E Net emf = E𝑒𝑞 = E Net resistance, 𝑅 + 𝑟 𝑛 Current in circuit, 𝑖 = E 𝑅 + 𝑟 𝑛
- 51. 𝑛 identical cells, each of emf 𝐸 and internal resistance 𝑟, are joined in series. Out of these, 𝑚 cells are wrongly connected, i.e., their terminals are connected in reverse of that required for series connection (𝑚 < 𝑛/2). Let 𝐸0 be the emf of the resulting battery and 𝑟o be its internal resistance, find 𝐸0, 𝑟o . 𝑟0 = 𝑛𝑟 So, if 𝑚 cells are connected wrongly among 𝑛 cells, Net emf, 𝐸0 = 𝑛 − 2𝑚 𝐸 • Since 𝑛 cells are connected in series, the internal resistances of the cells are also in series & since polarity doesn’t affect resistance. • Two oppositely connected cells connected in series nullify each other emfs. Therefore, this nullification always happens in pairs. 𝐴 𝐵 𝑖 E1 𝑟1 E2 𝑟2 E𝑁−𝑛 𝑟𝑛−𝑚 E1 𝑟1 E2 𝑟2 E𝑛 𝑟𝑚
- 52. 4 3 𝑉 2 3 Ω 𝐴 𝐵 3 𝑉 3 Ω 𝐴 𝐵 1 𝑉 2 𝑉 2 Ω 1 Ω 3 𝑉 3 Ω 𝐸′ 𝑟′ Calculate 𝐸𝑒𝑞 and 𝑟𝑒𝑞 for the given combination of cells between 𝐴 & 𝐵. 𝑟′ = 𝑟1𝑟2 𝑟1 + 𝑟2 𝐸′ = 𝐸1 𝑟1 + 𝐸2 𝑟2 1 𝑟1 + 1 𝑟2 ⇒ 𝐸′ = 4 3 𝑉 ⇒ 𝑟′ = 2 3 Ω 𝐸𝑒𝑞 = 𝐸′ + 3 𝐸𝑒𝑞 = 13 3 𝑉 ⇒ 𝑟𝑒𝑞 = 𝑟′ + 3 𝑟𝑒𝑞 = 11 3 Ω ⇒ = 4 3 + 3 = 2 3 + 3
- 53. Consider 𝑁 = 𝑛1𝑛2 identical cells, each of emf 𝐸 and internal resistance 𝑟. Suppose 𝑛1 cells are joined in series to form a line and 𝑛2 such lines are connected in parallel. The combination drives a current in an external resistance 𝑅. (a) Find the current through the external resistance. 1 𝑟𝑒𝑞 = 1 𝑛1𝑟 + 1 𝑛1𝑟 + ⋯ 1 𝑛1𝑟 = 𝑛2 𝑛1𝑟 ⇒ 𝑟𝑒𝑞= 𝑛1𝑟 𝑛2 𝐸𝑒𝑞 = 𝑛1𝐸 𝑛1𝑟 + 𝑛1𝐸 𝑛1𝑟 + ⋯ 𝑛1𝐸 𝑛1𝑟 1 𝑛1𝑟 + 1 𝑛1𝑟 + ⋯ 1 𝑛1𝑟 = 𝑛2𝐸 𝑟 𝑛2 𝑛1𝑟 ⇒ 𝐸𝑒𝑞 = 𝑛1𝐸 𝑖 = 𝑁𝐸 𝑛1𝑟 + 𝑛2𝑅 𝑖 = 𝑛1𝐸 𝑛1𝑟 𝑛2 + 𝑅
- 54. Current in the circuit, 𝑖 = 𝑁𝐸 𝑛1𝑟 + 𝑛2𝑅 = 𝑁𝐸 𝑛1𝑟 + 𝑁𝑅 𝑛1 Consider 𝑁 = 𝑛1𝑛2 identical cells, each of emf 𝐸 and internal resistance 𝑟. Suppose 𝑛1 cells are joined in series to form a line and 𝑛2 such lines are connected in parallel. The combination drives a current in an external resistance 𝑅. (b) Assuming that 𝑛1 and 𝑛2 can be continuously varied (𝑁 = 𝑛1𝑛2 still holds), find the relation between 𝑛1, 𝑛2, 𝑅 and 𝑟 for which the current in 𝑅 is maximum. When 𝑖 is maximum, 𝑖 = 𝑛1𝑁𝐸 𝑛1 2𝑟 + 𝑁𝑅 𝑑𝑖 𝑑𝑛1 = 0 𝑛1𝑟 = 𝑛2𝑅 ⇒ 𝑛1 2 𝑟 + 𝑁𝑅 − 2𝑛1 2 𝑟 = 0 ⇒ 𝑛1𝑛2𝑅 = 𝑛1 2 𝑟 (∵ 𝑁 = 𝑛1𝑛2) ⇒ ⇒ 𝑑𝑖 𝑑𝑛1 = 𝑁𝐸 𝑛1 2 𝑟 + 𝑁𝑅 − 𝑛1(𝑁𝐸)(2𝑛1𝑟) [𝑛1 2𝑟 + 𝑁𝑅]2 = 0 ∴ Condition for maximum current in 𝑅 is, 𝑛1𝑟 = 𝑛2𝑅
- 55. Heating Effect of Electric Current 𝑊 = 𝑉𝐼𝑡 = 𝐼2 𝑅𝑡 = 𝑉2 𝑅 𝑡 𝑗𝑜𝑢𝑙𝑒 • Taking 𝑉 in 𝑣𝑜𝑙𝑡, 𝐼 in 𝑎𝑚𝑝𝑒𝑟𝑒 and 𝑅 in 𝑜ℎ𝑚, • The work done appears as increased thermal energy of conductor and conductor gets heated. ∴ 𝐻 = 𝑉𝐼𝑡 = 𝐼2 𝑅𝑡 = 𝑉2 𝑅 𝑡 𝑗𝑜𝑢𝑙𝑒 𝑑𝑞- Charge passing through a potential difference, 𝑉 𝑑𝑞 𝑅 𝑉 When a charge moves through a potential difference, electrical work is done. This work represents the loss of potential energy of charges. 𝑑𝑊 = 𝑉𝑑𝑞 • Work done,
- 56. Power Power – rate at which energy is transferred For any electrical element, 𝑃 = 𝑉𝐼 𝑃- Instantaneous Power 𝑉- Potential difference across the element 𝐼- Instantaneous current through the element 𝐴 𝐵 𝐼 𝐸 • Battery acts as a source • Power is delivered 𝐴 𝐵 𝐸 𝐼 • Battery acts as a load • Power is absorbed 𝐴 𝐵 𝐸 𝐼
- 57. Power Dissipation Across a Resistor • A resistance always acts as a load. It is the rate of generation of thermal energy across the resistor. 𝑅 𝐴 𝐵 𝑖 𝐸 𝑃𝑅 = 𝑉𝑖 = 𝑖𝑅 𝑖 = 𝑖2 𝑅 = 𝑉 𝑉 𝑅 = 𝑉2 𝑅 𝑃 = 𝑖2 𝑅 = 𝑉𝑖 = 𝑉2 𝑅
- 58. Maximum Power Dissipation Across a Resistor 𝑃𝑅 + 𝑃𝑟 = 𝐸2 𝑅 + 𝑟 ⇒ 𝑃𝑅 + 𝑃𝑟 = 𝑃𝐸 𝑅 𝐴 𝐵 𝑖 𝐸 𝑟 • 𝑃𝑅 = 𝑖2 𝑅 =. 𝐸2 𝑅 (𝑅 + 𝑟)2 • 𝑖 =. 𝐸 𝑅 + 𝑟 • 𝑃𝑟 = 𝑖2 𝑟 =. 𝐸2 𝑟 (𝑅 + 𝑟)2 • 𝑃𝐸 = 𝐸𝑖 =. 𝐸2 𝑅 + 𝑟 𝑃𝑚𝑎𝑥 𝑅 = 𝐸2 4𝑟 for 𝑅 = 𝑟 𝑃𝑅 = 𝐸2 𝑅 (𝑅 + 𝑟)2 When 𝑃𝑅 is maximum, 𝑑𝑃𝑅 𝑑𝑅 = 0 ⇒ 𝐸2 𝑅 + 𝑟 2 − 2 𝑅 + 𝑟 . 𝐸2 𝑅 (𝑅 + 𝑟)4 = 0 ⇒ 𝐸2 𝑅 + 𝑟 𝑅 + 𝑟 − 2𝑅 = 0 ⇒ 𝑅 = 𝑟
- 59. A battery has an open circuit potential difference of 6 𝑉 between its terminals. When a load resistance of 60 Ω is connected across the battery, the total power supplied by the battery is 0.4 𝑊. What should be the load resistance 𝑅, so that maximum power will be dissipated in 𝑅. Calculate this power. What is the total power supplied by the battery when such a load is connected? 𝑅 6 𝑉 𝑟 𝑖 𝑅 = 30 Ω 𝑖 = 6 60 + 𝑟 𝑃𝑏𝑎𝑡𝑡𝑒𝑟𝑦 = 6𝑖 = 36 60 + 𝑟 ⇒ 36 60 + 𝑟 = 0.4 = 2 5 ⇒ 𝑟 = 30 Ω For maximum 𝑃𝑅, 𝑅 = 𝑟 𝑃 𝑚𝑎𝑥 𝑅 = 𝐸2 4𝑟 = 62 4 × 30 ⇒ 𝑃𝑚𝑎𝑥 𝑅 = 0.3 𝑊 𝑃𝐸 = 0.6 𝑊 𝑃𝐸 = 𝑃 𝑚𝑎𝑥 𝑅 + 𝑃𝑟 𝑃𝑟 = 𝑃 𝑚𝑎𝑥 𝑅 (∵ 𝑅 = 𝑟) ⇒ 𝑃𝑟 = 0.3 𝑊
- 60. Measuring Electric Energy 1 𝑘𝑊ℎ = 𝑃(𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡) × 𝑡(ℎ𝑜𝑢𝑟) 1 𝑘𝑊ℎ = 1000 × 3600 𝑗𝑜𝑢𝑙𝑒 1 𝑘𝑊ℎ = 3.6 × 106 𝑗𝑜𝑢𝑙𝑒 • SI unit of energy is joule( 𝐽) • 𝑘𝑊ℎ is a larger unit 𝑅 𝐴 𝐵 𝑖 𝐸 (𝑁) = 𝑤𝑎𝑡𝑡 × ℎ𝑜𝑢𝑟 1000 OR • No. of units 𝑁 = Power consumed in watts Time of consumption in hours 1000
- 61. 𝑅 = 𝑉2 𝑃 Resistance of an Electric Bulb 𝑉 𝑃, 𝑉 ≡ 𝑅 𝑉 Bulb Rating: 𝑃 - Power 𝑉 - Potential difference Bulb consumes 𝑃 power for an applied potential difference of 𝑉. Resistance of the bulb, 𝑖 𝑖
- 62. How will the Bulbs Glow? 𝑅 = 𝑉2 𝑃 𝑅 = 2202 60 = 1502 × 60 2202 ⇒ 𝑃1 < 60 𝑊 𝑃1 = 𝑉1 2 𝑅 = 4002 × 60 2202 ⇒ 𝑃3 > 60 𝑊 = 2202 × 60 2202 ⇒ 𝑃2 = 60 𝑊 𝑃2 = 𝑉2 2 𝑅 𝑃3 = 𝑉3 2 𝑅 • Bulb will glow dimmer • Bulb will glow with its rated power • Bulb will fuse 𝑖 60 𝑊, 220 𝑉 150 𝑉 𝑖 60 𝑊, 220 𝑉 220 𝑉 𝑖 60 𝑊, 220 𝑉 400 𝑉
- 63. 𝑃2 = 4 𝑊 𝑃1 = 16 𝑊 𝑅1 = 𝑉1 2 𝑃 = 2202 25 𝑅2 = 𝑉2 2 𝑃 = 2202 100 (𝑎) 𝑅1 = 4𝑅2 ⟹ 𝑉1 = 4𝑉2 ∴ 𝑉1 = 4 5 × 220 𝑃1 = 𝑉1 2 𝑅1 𝑃2 = 𝑉2 2 𝑅2 = 16 × 2202 25 × 25 2202 = 2202 25 × 100 2202 Bulb 𝐵1 glows more brightly Two electric bulbs 𝐵1 and 𝐵2 rated at 25 𝑊, 220 𝑉 and 100 𝑊, 220 𝑉 are given. Which bulb glows more brightly and what is the value of output powers 𝑃1 and 𝑃2 , if (𝑎) they are connected in series across a 220 𝑉 battery (𝑏) connected in parallel across a 220 𝑉 battery 𝑉2 = 1 5 × 220 (𝑏) 𝑃1 = 25 𝑊, 𝑃2 = 100 𝑊 Applied voltage = Rated voltage Bulb 𝐵2 glows more brightly
- 64. A 100 𝑊, 250 𝑉 bulb 𝐵1, and two 60 𝑊, 250 𝑉 bulbs 𝐵2 and 𝐵3, are connected to a 250 𝑉 source as shown. Now 𝑊1, 𝑊2 and 𝑊3 are the output powers of bulbs 𝐵1, 𝐵2 and 𝐵3. Then find relation between 𝑊1, 𝑊2 and 𝑊3 . Resistance provided by the bulbs, 𝑅1 = 250 2 100 Ω = 𝑅 say 𝑅3 = 𝑅2 = 250 2 60 Ω = 5𝑅 3 For the given series connection, 𝑉 + 5𝑉 3 = 250 ⇒ 𝑉 = 3 8 250 𝑉 ⇒ 𝑉1 = 𝑉 = 3 8 250 𝑉 ⇒ 𝑉2 = 5𝑉 3 = 5 8 250 𝑉 𝐵1 𝐵2 +250 𝑉 𝑅, 𝑉 0 𝑉 5𝑅 3 , 5𝑉 3
- 65. Output powers of bulbs 𝐵1 and 𝐵2, 𝑊1 = 𝑉1 2 𝑅1 = 9 64 250 2 × 100 250 2 = 14.06 𝑊 𝐵1 𝐵2 +250 𝑉 𝑅, 𝑉 0 𝑉 5𝑅 3 , 5𝑉 3 𝑊2 = 𝑉2 2 𝑅2 = 25 64 250 2 × 60 250 2 = 23.44 𝑊 ⇒ 𝑉1 = 𝑉 = 3 8 250 𝑉 ⇒ 𝑉2 = 5𝑉 3 = 5 8 250 𝑉 Since 𝐵3 is at its rated voltage, 𝑊3 = 60 𝑊 𝑊1 < 𝑊2 < 𝑊3
- 66. Electrical Measuring Instruments Galvanometer Ammeter • Detects the presence of electric current in a circuit. • Reading of Galvanometer ∝ 𝑖 • Current measured by the galvanometer is, 𝑖′ = E 𝑅 + 𝑅𝑔 𝑅𝑔 ≈ 0 for accurate reading • Measures current passing through it • Minimal effect on existing current in the circuit (negligible resistance) • Connected in series in the circuit 𝑖′ = E 𝑅 + 𝑅𝑎 𝑅𝑎 ≈ 0 for an ideal ammeter
- 67. Electrical Measuring Instruments Voltmeter Potentiometer • Measures potential difference across it • Connected in parallel with the circuit • Resistance of an ideal voltmeter is infinite so that no current passes through it • Measures potential difference without drawing any current from the given circuit. • Jockey is moved from one end to the other end of the wire AB to find the null point (zero deflection) 𝑉 𝑎 − 𝑉𝑏 = 𝑉𝐴 − 𝑉𝑃 = 𝑍𝑙 Where, 𝑍 = Potential gradient = 𝑉𝐴 − 𝑉𝐵 𝐿 𝑙
- 68. Conversion of Galvanometer into Ammeter • Full-scale deflection current 𝑖𝑔 is the maximum current that is allowed to pass through the galvanometer. • A very small resistance 𝑟𝑆 (shunt) is connected in parallel to minimize the resistance of the ammeter. 𝑉 𝑔 = 𝑉𝑆 𝑖𝑔𝑅𝑔 = 𝑖 − 𝑖𝑔 𝑟𝑆 , where 𝑖 is the range of the ammeter. Ammeter 𝑖 𝑟𝑆 G 𝑖𝑔 𝑖 − 𝑖𝑔 𝑅𝑔 A 𝑖 𝑟𝑆 = 𝑖𝑔𝑅𝑔 𝑖 − 𝑖𝑔
- 69. A galvanometer has resistance 2 Ω with maximum current measuring capacity 0.1 𝐴. Find the shunt resistance 𝑆 of an ammeter which can measure current up to 10 𝐴. Maximum current allowed inside galvanometer is, 𝑖𝑔 = 0.1 𝐴 Shunt 𝑆 is such that galvanometer can measure up to 10 𝐴 current flowing through the circuit. 𝑖 = 10 𝐴 𝑅 = 2 Ω G 𝑆 𝑟 𝑔 = 2 Ω 𝑖𝑔 𝑖 − 𝑖𝑔 Applying KVL inside the loop containing 𝐺 and 𝑆, 𝑟 𝑔𝑖𝑔 = 𝑖 − 𝑖𝑔 𝑆 2 × 0.1 = 10 − 0.1 × 𝑆 0.2 = 9.9𝑆 𝑆 = 2 99 Ω
- 70. Conversion of Galvanometer into Voltmeter • Max potential difference that can be measured with voltmeter is, 𝑉𝐴𝐵 = 𝑖𝑔 𝑅𝑔 + 𝑅 ≈ 𝑖𝑔𝑅 ∵ 𝑅𝑔 ≪ 𝑅 • If the reading of the newly calibrated voltmeter is 𝑉, then, 𝑉 = 𝑖𝑔 𝑅𝑔 + 𝑅 𝑅′ 𝑖𝑔 𝑖 − 𝑖𝑔 𝑅𝑔 𝑖 𝑅 Voltmeter 𝐴 𝐵 𝑅 = 𝑉 𝑖𝑔 − 𝑅𝑔
- 71. A galvanometer has a coil of resistance 100 Ω showing a full-scale deflection at 50 𝜇𝐴. What resistance should be added to use it as a voltmeter of range 50 𝑉? Applying KVL across the circuit, 𝑖𝑔 𝑅𝑔 + 𝑅 = 𝑉 ⇒ 𝑅 = 𝑉 𝑖𝑔 − 𝑅𝑔 = 50 50 × 10−6 − 100 𝑅 ≈ 106 Ω 𝑅𝑔 = 100 Ω 𝑅 𝑖𝑔 = 50 𝜇𝐴 G 𝑉 = 50 𝑉
- 72. Calibration of Potentiometer where, • If the null point is found at point 𝑃 which is at a distance 𝑙 from point 𝐴, then, 𝑉𝐴 − 𝑉𝑃 = E′ = 𝑉𝑂 𝐿 𝑙, 𝑉𝑂 = 𝑉𝐴 − 𝑉𝐵 = E𝑅 𝑅 + 𝑟 𝑅 = Resistance of the potentiometer wire E1 E2 = 𝑙1 𝑙2
- 73. Measurement of Internal Resistance • When the key is open, no current flows through the secondary circuit. If the null point is found at point 𝑃 (length 𝑙1), then, E1 = 𝑉𝐴𝐵 𝑙1 𝐿 • At null deflection (𝑃2), ⇒ 𝑉𝐴𝐵 𝑙2 𝐿 = 𝑖′𝑅 ⇒ 𝑉𝐴𝐵 𝑙2 𝐿 = E1𝑅 𝑅 + 𝑟 𝑉𝐴 − 𝑉𝑃2 = 𝑉 𝑎 − 𝑉𝑏 E1 = 𝑉𝐴𝐵 𝑙1 𝐿 ; E1𝑅 𝑅 + 𝑟 = 𝑉𝐴𝐵 𝑙2 𝐿 By dividing these two, = 1 + 𝑟 𝑅 𝑙1 𝑙2 = 𝑅 + 𝑟 𝑅 𝑟 = 𝑅 𝑙1 − 𝑙2 𝑙2
- 74. A 6 𝑉 battery of negligible internal resistance is connected across a uniform wire 𝐴𝐵 of length 100 𝑐𝑚. The positive terminal of another battery of emf 4 𝑉 and internal resistance 1 Ω is joined to the point 𝐴 as shown. Take the potential at 𝐵 to be zero. (a) What are the potentials at 𝐴 and 𝐶? (b) At what point 𝐷 of the wire 𝐴𝐵, the potential is equal to that at 𝐶? (c) If points 𝐶 and 𝐷 are connected by a wire, what will be the current through it? Extrapolating potential at various points, 𝑉𝐴 = 6 𝑉, 𝑉𝐶 = 2 𝑉 Thus, the voltages at 𝐴 and 𝐶 are, 𝐴𝐷 = 66.67 𝑐𝑚 Taking the potential gradient, Given, 𝑉𝐷 = 𝑉𝐶 Also, 𝑉𝐴 = 6 𝑉 𝑉𝐶 = 2 𝑉 𝐴𝐵 = 𝐿 = 100 𝑐𝑚 𝑉𝐴 − 𝑉𝐵 𝐿 = 𝑉𝐴 − 𝑉𝐷 𝑙 Since, 𝑉𝐴 − 𝑉𝐵 = 6 𝑉 and 𝑉𝐴 − 𝑉𝐷 = 4 𝑉 We get, 𝑙 = 2 3 𝐿 Thus, current across 𝐷𝐶, 𝑖 = 0 𝑖 = 𝑉𝐷 − 𝑉 𝑐 𝑅
- 75. Charging a RC Circuit • At time 𝑡, the charge on the capacitor is 𝑞 and current 𝑖 • At time 𝑡 = 0, the charge on the capacitor is 0 and current 𝑖 = 𝑖0 = E 𝑅 𝑅 E 𝐶 At 𝑡 = 0 E E E 0 0 0 𝑅 E At 𝑡 = 0+ 𝑖𝑜 𝑅 E 𝐶 At 𝑡 𝑞 𝐶 𝑖𝑅 𝑞 𝑖 + − 𝑅2 𝐶 𝑅 𝐶 E 𝑅2 An RC circuit is a circuit defined by the combination of a resistor and a capacitor
- 76. Derivation of Charging RC Circuit Applying KVL At time 𝑡, − 𝑞 𝐶 − 𝑖𝑅 + E = 0 𝑖 = 𝑑𝑞 𝑑𝑡 = 𝐶E − 𝑞 𝑅𝐶 Thus, we get, Integrating using variable separable method, න 0 𝑞 𝑑𝑞 𝐶E − 𝑞 = 1 𝑅𝐶 න 0 𝑡 𝑑𝑡 ln 𝐶E − 𝑞 −1 𝑜 𝑞 = 1 𝑅𝐶 𝑡 𝑜 𝑡 Applying the limits, Thus, we get, Differentiating w.r.t time, we get current as, ln 𝐶E − 𝑞 𝐶E = − 𝑡 𝑅𝐶 𝑞 = 𝐶E 1 − 𝑒− 𝑡 𝑅𝐶 𝑖 = E 𝑅 𝑒− 𝑡 𝑅𝐶
- 77. Derivation of Charging RC Circuit At 𝑡 = ∞, 𝑅 E 𝐶 At 𝑡 = ∞ E E 0 0 E 0 • As time goes on, the charge on the capacitor increases. • As time goes on, current across the circuit decreases. (Steady State) 𝑞 = 𝐶E ; 𝑖 = 0 𝑖 𝑡 𝑖0 = E 𝑅 𝑞 𝑡 𝐶E 𝑞 = 𝐶E 1 − 𝑒− 𝑡 𝑅𝐶 𝑖 = E 𝑅 𝑒− 𝑡 𝑅𝐶
- 78. Time Constant of a RC Circuit 𝑅 E 𝐶 𝑞 𝑖 The product 𝑅𝐶 is called the time constant (𝜏) and its SI unit is seconds. Thus, at 𝑡 = 𝜏, we get, Similarly, at 𝑡 = 𝜏 , we get current as, 𝑞 = 𝐶E 1 − 𝑒−1 ≈ 0.63𝐶E 𝑖 = E 𝑅 𝑒− 1 = 0.37 E 𝑅 𝑅𝐶 = 𝜏 + − 𝒕 𝒒 𝒊 𝑡 𝐶E E 𝑅 𝑒− 𝑡 𝑅𝐶 0 0 E 𝑅 𝜏 0.63𝐶E 0.37 E 𝑅 ∞ 𝐶E 0 Values of charge and current at different time instances are as follows, 1 − 𝑒− 𝑡 𝑅𝐶
- 79. Time Constant of an RC Circuit • Nearly 95% of the charging occurs within three time constants. • It takes five time constants to reach almost (99%) steady state. 𝑅 E 𝐶 𝑞 𝑖 𝑞 𝑡 0.99𝐶E 0.95𝐶E 0.86𝐶E 0.63𝐶E 𝜏 2𝜏 3𝜏 5𝜏 𝑞 = 𝐶E 1 − 𝑒− 𝑡 𝑅𝐶 + −
- 80. What is the steady state current in the 2 Ω resistor shown in the circuit? The internal resistance of battery is negligible and the capacitance of condenser 𝐶 is 0.2 𝜇𝐹. At steady state, current across the condenser is zero and behaves as an open circuit (branch) 𝑖2Ω = 0.9 𝐴 𝑖𝑜 = 6 𝑅𝑒𝑞 = 6 2.8 + 1 1 2 + 1 3 = 6 2.8 + 1.2 = 6 4 = 1.5 𝐴 By ratio of currents across parallel resistors, 3𝑖 + 2𝑖 = 1.5 𝐴 ⇒ 𝑖 = 0.3 𝐴 Thus, current across 2 Ω resistor = 3𝑖 Thus, at 𝑡 = ∞, we get,
- 81. • At time 𝑡 = 0, the p.d. across the capacitor is, 𝑉 = 𝑄 𝐶 • Just after the circuit is completed, an initially charged capacitor behaves as a battery having EMF 𝑄 𝐶 and the current through the resistor is, 𝑖0 = 𝑄 𝑅𝐶 𝑖0 𝑅 𝐶, + − 𝑄 𝐶 Discharging RC Circuit • The relation between 𝑖 and 𝑞 is, 𝑖 = − 𝑑𝑞 𝑑𝑡 The – 𝑣𝑒 sign indicates that the charge in the capacitor is decreasing as time progresses. • Apply KVL in the circuit at time 𝑡: − 𝑞 𝐶 + 𝑖𝑅 = 0 𝑖 = 𝑞 𝑅𝐶 = − 𝑑𝑞 𝑑𝑡 𝑑𝑞 𝑞 = − 𝑑𝑡 𝑅𝐶 𝑞 + − 𝑖 𝑅, 𝑖𝑅 − + 𝐶, 𝑄 𝐶
- 82. Derivation of Discharging RC Circuit Integrating, we get, න 𝑄 𝑞 𝑑𝑞 𝑞 = − 1 𝑅𝐶 න 0 𝑡 𝑑𝑡 After solving we get, 𝑞 = 𝑄𝑒− 𝑡 𝑅𝐶 Differentiating w.r.t time, we get current as, 𝑖 = Q 𝑅𝐶 𝑒− 𝑡 𝑅𝐶 At 𝑡 = ∞, • As time goes on, the charge on the capacitor decreases. • As time goes on, current across the circuit decreases. (Steady State) 𝑞 = 𝐶E ; 𝑖 = 0 𝑖 𝑡 Q 𝑅𝐶 𝑞 𝑡 𝑄 𝑞 + − 𝑖 𝑅, 𝑖𝑅 − + 𝐶, 𝑄 𝐶
- 83. • Nearly 95% of the discharging occurs within 3 times constants. • It takes 5 times constants, to reach almost (99% discharging) steady state. 𝑞 𝑡 0.01𝑄 0.05𝑄 0.14𝑄 0.37𝑄 𝜏 2𝜏 3𝜏 5𝜏 𝑄 𝑞 + − 𝑖 𝐶 𝑅 𝑞 𝐶 𝑖𝑅 − + 𝒕 𝒒 𝒊 𝑡 𝑄𝑒− 𝑡 𝑅𝐶 Q 𝑅𝐶 𝑒− 𝑡 𝑅𝐶 0 𝑄 Q 𝑅𝐶 𝜏 0.37𝑄 0.37 𝑄 𝑅𝐶 ∞ 0 0 Discharging RC Circuit
- 84. How many time constants will elapse before the charge on a capacitor falls to 0.1% of its maximum value in a discharging RC circuit? We know, Let 𝑡 = 𝑛𝜏. Thus, at 𝑡, 𝑛 ≈ 6.9 𝑛 = 3 ln 10 To find 𝑛, 0.5 𝐹 2.8 Ω 𝑄 1000 = 𝑄𝑒− 𝑛𝜏 𝜏 1 1000 = 𝑒−𝑛 ⇒ 𝑒𝑛 = 1000 ∴ ,i.e., Time taken is 6.9 𝜏 𝑞 = 𝑄𝑒− 𝑡 𝜏 𝑖 = Q 𝑅𝐶 𝑒− 𝑡 𝜏 = 𝑖0𝑒− 𝑡 𝜏 𝑞 = 0.1 100 𝑄 = 𝑄 1000
- 85. Find the time constant for the given 𝑅𝐶 circuit. [Given: 𝑅1 = 1 Ω, 𝑅2 = 2 Ω, 𝐶1 = 4 𝜇𝐹, 𝐶1 = 2 𝜇𝐹] • Time constant for 𝑅𝐶 circuit : • Equivalent resistance : • Equivalent capacitance : 𝜏 = 𝑅𝑒𝑞𝐶𝑒𝑞 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 𝑅𝑒𝑞 = 1 Ω + 2 Ω = 3 Ω 𝐶𝑒𝑞 = 𝐶1 + 𝐶2 𝐶𝑒𝑞 = 4 𝜇𝐹 + 2 𝜇𝐹 = 6 𝜇𝐹 (Resistors are in series) (Capacitors are in parallel connection) • Time constant: 𝜏 = 𝑅𝑒𝑞𝐶𝑒𝑞 = 3 × 6 = 18 𝜇𝑠 𝜏 = 18 𝜇𝑠 𝑖 1 Ω 2 Ω 4 𝜇𝐹 2 𝜇𝐹 E
- 86. At 𝑡 = 0, switch 𝑆 is closed. The charge on the capacitor is varying with time as 𝑄 = 𝑄0 1 − 𝑒−𝛼𝑡 . Obtain the value of 𝑄0 and 𝛼 in terms of given circuit parameters. • For charging RC circuit : 𝑄 = 𝑄0 1 − 𝑒− 𝑡 𝜏 𝑄0 = Charge stored in the capacitor at steady state = 𝐶 × P.D. across the capacitor at steady state = 𝐶 × P.D. across the resistance 𝑅2 = 𝐶 × 𝑅2 × Current through the resistance 𝑅2 = 𝐶𝑅2 × 𝑉 𝑅1 + 𝑅2 𝑄0 = 𝐶𝑉𝑅2 𝑅1 + 𝑅2
- 87. • Given : 𝑄 = 𝑄0 1 − 𝑒− 𝛼𝑡 𝛼 = 𝑅1 + 𝑅2 𝐶𝑅1𝑅2 • We have : 𝑄 = 𝑄0 1 − 𝑒− ൗ 𝑡 𝜏 𝛼 = 1 𝜏 • Time constant of the given circuit : 𝜏 = 𝐶𝑅𝑒𝑞 𝑅𝑒𝑞 = Equivalent resistance across the capacitor after short circuiting the battery. 𝑅1 𝑅2 𝑅𝑒𝑞 = 𝑅1𝑅2 𝑅1 + 𝑅2 • Finally, 𝜏 = 𝐶𝑅𝑒𝑞 = 𝐶𝑅1𝑅2 𝑅1 + 𝑅2 𝛼 = 1 𝜏
- 88. A capacitor is charged using an external battery with a resistance 𝑥 in series. The dashed line shows the variation of ln 𝐼 with respect to time. If the resistance is changed to 2𝑥, the new graph will be 𝑃 𝑄 𝑅 𝑆 ln 𝐼 𝑡 • For charging RC circuit : 𝐼 = E 𝑅 𝑒−𝑡/𝑅𝐶 • Taking ln on both sides: ln 𝐼 = ln E 𝑅 − 𝑡 𝑅𝐶 • Resistance is increased from 𝑥 to 2𝑥 Slope will become less steeper Slope = − 1 𝑅𝐶 • At 𝑡 = 0, we have : ln 𝐼 = ln E 𝑅 As resistance is increased, ln 𝐼 will be decreased. Graph 𝑄 represents the correct behaviour
- 89. How many time constants will elapse before the energy stored in a capacitor reaches half of its equilibrium value in a charging 𝑅𝐶 circuit. • For charging RC circuit : 𝑞 = 𝐶E 1 − 𝑒− 𝑡 𝑅𝐶 = 𝑄 1 − 𝑒− 𝑡 𝜏 • The energy stored in the capacitor at equilibrium, 𝑈0 = 𝑄2 2𝐶 𝑈 = 𝑞2 2𝐶 = 𝑈0 2 𝑞2 2𝐶 = 𝑄2 4𝐶 𝑞 = 𝑄 2 • Assume that energy stored in the capacitor goes from 𝑈0 to 𝑈0 2 in time 𝑡 = 𝑛𝜏. • At 𝑡 = 𝑛𝜏, energy stored in the capacitor, 𝑖 𝑞 + − 𝑅 E
- 90. • Finally, 𝑞 = 𝑄 1 − 𝑒− 𝑡 𝜏 𝑒−𝑛 = 1 − 1 2 𝑄 2 = 𝑄 1 − 𝑒− 𝑛𝜏 𝜏 𝑒𝑛 = 2 2 − 1 = 2 2 + 1 𝑛 = ln 2 + 2 = 1.23 𝑡 = 1.23𝜏 𝑖 𝑞 + − 𝑅 E 𝐶
- 91. Find the potential difference between the points 𝐴 and 𝐵 and between the points 𝐵 and 𝐶 at the steady state. 0 0 0 100 100 100 100 0 𝑉 • At steady state, capacitors behave as an open circuit. • Choose zero potential at any specific junction and find potential of other junctions with respect to that. • Now, if any particular junction is left out, choose its potential to be 𝑉.
- 92. • Apply KCL (Conservation of charge) in the annotated island: 3 𝑉 − 100 + 3 𝑉 − 100 + 1 𝑉 − 0 + 1 𝑉 − 0 = 0 𝑉 = 75 𝑉𝑜𝑙𝑡𝑠 • Potential difference between any two junctions of a circuit should be positive just like difference between any two natural numbers is positive. • Potential difference between 𝐴 and 𝐵 : 𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 = 25 𝑉𝑜𝑙𝑡𝑠 • Potential difference between 𝐵 and 𝐶 : 𝑉𝐵𝐶 = 𝑉𝐵 − 𝑉𝐶 = 75 𝑉𝑜𝑙𝑡𝑠
- 93. A capacitor of capacitance 𝐶 charged to a potential difference 𝑉 is discharged by connecting its plates through a resistance 𝑅. Find the heat dissipated in one time constant after the connections are made. Do this by calculating ∫ 𝑖2 𝑅 𝑑𝑡 and also by finding the decrease in the energy stored in capacitor. 𝑖 𝐶 𝑅 • For discharging RC circuit : 𝑖 = 𝑄 𝑅𝐶 𝑒− 𝑡 𝑅𝐶 = 𝑖0𝑒− 𝑡 𝜏 • Heat dissipated in one time constant : 𝐻 = න 0 𝜏 𝑖2 𝑅 𝑑𝑡 = න 0 𝜏 𝑖0 2 𝑅 𝑒− 2𝑡 𝜏 𝑑𝑡 𝐻 = 𝑄2 2𝐶 1 − 1 𝑒2 Or 𝐻 = 1 2 𝐶𝑉2 1 − 1 𝑒2
- 94. 𝑖 𝐶 𝑅 and also by finding the decrease in the energy stored in capacitor. • Initial charge on the capacitor = 𝑄 Initially, energy stored in the capacitor, 𝑈0 = 𝑄2 2𝐶 • After one time constant, Charge remaining on the capacitor, 𝑞 = 0.37𝑄 = 𝑄 𝑒 Energy stored in the capacitor, 𝑈𝜏 = 𝑞2 2𝐶 = ൗ 𝑄2 2𝐶 𝑒2 • Energy lost, 𝑈𝑙𝑜𝑠𝑡 = 𝑈0 − 𝑈𝜏 = 𝑄2 2𝐶 1 − 1 𝑒2 = 𝐻
- 95. At 𝑡 = 0, 𝑞 = 𝑄 At 𝑡 = 0, 𝑞1 = 0 𝑞1(𝑡) = 𝐶𝐸(1 − 𝑒−𝑡/𝑅𝐶 ) At 𝑡 = 0, 𝑞2 = 𝑄 𝑞2(𝑡) = 𝑄(𝑒−𝑡/𝑅𝐶 ) A capacitor of capacitance 𝐶 is given a charge 𝑄. At 𝑡 = 0, it is connected to an ideal battery of emf 𝐸 through a resistance 𝑅. Find the charge on the capacitor at time 𝑡. ≡ + 𝑖 𝑞 + − 𝑅 𝐸 𝐶 𝑅 𝐸 𝐶 𝑞2 + − 𝑅 𝐶 𝑞1 + − ⇒ 𝑞 𝑡 = 𝐶𝐸 1 − 𝑒− 𝑡 𝑅𝐶 + 𝑄 𝑒− 𝑡 𝑅𝐶 𝑞 𝑡 = 𝑞1 𝑡 + 𝑞2(𝑡)
- 96. −𝑖𝑐𝑅 − 𝑞 𝐶 − 𝑖𝑅 + 𝐸 = 0 ⇒ 𝐸 = 𝑖𝑅 + 𝑖𝑐𝑅 + 𝑞 𝐶 Outer loop: Loop 1: − 𝑖 − 𝑖𝑐 𝑅 − 𝑖𝑅 + 𝐸 = 0 ⇒ 𝐸 = 2𝑖𝑅 − 𝑖𝑐𝑅 1 2 Putting value of 𝑖𝑅 from 2 in 1 , 𝐸 = 𝐸 + 𝑖𝐶𝑅 2 + 𝑖𝑐𝑅 + 𝑞 𝐶 In the circuit shown in the figure, the battery is an ideal one, with emf 𝐸, the capacitor is initially uncharged. The switch is closed at 𝑡 = 0. (𝑎) Find the charge on the capacitor at time 𝑡. ⇒ 𝑖𝑐 = 𝐶𝐸 − 2𝑞 3𝑅𝐶 ⇒ 𝑑𝑞 𝑑𝑡 = 𝐶𝐸 − 2𝑞 3𝑅𝐶 ⇒ 𝑑𝑞 𝐶𝐸 − 2𝑞 = 𝑑𝑡 3𝑅𝐶 ⇒ න 0 𝑞 𝑑𝑞 𝐶𝐸 − 2𝑞 = 1 3𝑅𝐶 න 0 𝑡 𝑑𝑡 ⇒ ln(𝐶𝐸 − 2𝑞) −2 0 𝑞 = 1 3𝑅𝐶 𝑡 0 𝑡 ⇒ 𝑞 = 𝐶𝐸 2 1 − 𝑒− 2𝑡 3𝑅𝐶 Integrating,
- 97. 𝑞 = 𝐶𝐸 2 1 − 𝑒− 2𝑡 3𝑅𝐶 𝑖𝐶 = 𝑑𝑞 𝑑𝑡 = 𝐸 3𝑅 𝑒− 2𝑡 3𝑅𝐶 −𝑖𝑐𝑅 − 𝑞 𝐶 − 𝑖𝑅 + 𝐸 = 0 ⇒ 𝑖 = 𝐸 𝑅 − 𝑖𝐶 − 𝑞 𝑅𝐶 ⇒ 𝑖 = 𝐸 𝑅 − 𝐸 3𝑅 𝑒− 2𝑡 3𝑅𝐶 − 1 𝑅𝐶 𝐶𝐸 2 1 − 𝑒− 2𝑡 3𝑅𝐶 ⇒ 𝑖 = E 2𝑅 + E 6𝑅 𝑒− 2𝑡 3𝑅𝐶 At 𝑡 = ∞: 𝑖 = E 2𝑅 𝑅 𝐸 𝐶 𝑅 𝑅 𝐵 𝑆 𝑖 (𝑖 − 𝑖𝑐) 𝑖𝑐 𝐴 + 𝑞 − Apply KVL in the outer loop: In the circuit shown in the figure, the battery is an ideal one, with emf 𝐸, the capacitor is initially uncharged. The switch is closed at 𝑡 = 0. (𝑏) Find the current in 𝐴𝐵 at time 𝑡. What is its limiting value as 𝑡 → ∞.
- 98. = 𝑉𝐴𝐵 = 𝐸 2 Calculate 𝑉 ∞: 𝑞 = 𝐶𝑉 ∞(1 − 𝑒 − 𝑡 𝜏𝑒𝑓𝑓 ) ⇒ 𝑞 = 𝐶𝐸 2 (1 − 𝑒− 2𝑡 3𝑅𝐶) Charge on the capacitor: 𝑉 ∞ = Steady state voltage across C ⇒ 𝑉 ∞ = 𝐸 2 In the circuit shown in the figure, the battery is an ideal one, with emf 𝐸, the capacitor is initially uncharged. The switch is closed at 𝑡 = 0. (𝑎) Find the charge on the capacitor at time 𝑡. 𝜏𝑒𝑓𝑓 = 𝐶𝑅𝑒𝑞 𝑅𝑒𝑞 = 𝑅𝐴𝐵 ∴ 𝜏𝑒𝑓𝑓 = 3𝑅𝐶 2 = 𝑅 × 𝑅 𝑅 + 𝑅 + 𝑅 = 3𝑅 2 ⇒ 𝑅𝑒𝑞 = 3𝑅 2 Calculate 𝜏𝑒𝑞: Alternate Solution :
- 99. In the given circuit, 𝐶 = 5 𝜇𝐹. Find the current in 𝑅3 and the energy stored in the capacitor. Current in the 4Ω, 𝑖 = 6 − 0 4 ⇒ 𝑖 = 1.5 𝐴 Applying junction rule at A, 𝑉 − 2 − 0 2 + 𝑉 + 3 − 6 3 = 0 ⇒ 𝑉 = 12 5 𝑉 Energy in stored in the capacitor, 𝐸 = 1 2 𝐶𝑉2 ⇒ 𝐸 = 1.44 × 10−5 𝐽 = 1 2 × 5 × 10−6 × 12 5 2
- 100. Part of a circuit in a steady state along with the currents flowing in the branches, is shown in the figure. Calculate the energy stored in the capacitor. Current across the capacitor in steady state, Apply KCL at 𝑎: 2 + 1 + 0 = 𝐼1 ⇒ 𝐼1 = 3 𝐴 Apply KCL at 𝑏: −2 + 1 + 𝐼2 = 0 ⇒ 𝐼2 = 1 𝐴 𝐼𝐶 = 0 𝐴 𝑉 𝑎 − 5𝐼1 − 𝐼1 − 2𝐼2 − 𝑉𝑏 = 0 𝑉 𝑎 − 𝑉𝑏 = 6𝐼1 + 2𝐼2 = 6 × 3 + 2 × 1 = 20 𝑉 Apply KVL in the loop 1: Energy stored in the capacitor: 𝐸 = 1 2 𝐶𝑉2 = 1 2 × 4 × 10−6 × 202 ⇒ 𝐸 = 8 × 10−4 𝐽