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PHXII03:CURRENT ELECTRICITY

357482 A network of resistances, cell and capacitor $C (= 2 μ F)$ is shown in figure. In steady state condition, the charge on $2 μ F$ capacitor is $Q$ , while $R$ is unknown resistance. Values of $Q$ and $R$ are respectively
supporting img

1

4 μ C

and

10 Ω

2

4 μ C

and

4 Ω

3

2 μ C

and

2 Ω

4

8 μ C

and

4 Ω

Explanation:

In the steady state, capacitor acts as a open circuit, hence no current will flow through $4 Ω$ .
$∴$ The currents distribution is shown in figure.
supporting img
The potential difference between $C$ and $D$ is
$V_{C} - V_{D} = (1 A) (2 Ω) = 2 V$
As $V_{D} = 0 V$ ( $D$ is earthed)
$∴ V_{C} = 2 V$
The potential drop across the resistance $R$
$= 10 V + 2 V - 2 V = 10 V$
$∴$ Resistance, $R = \frac{10 V}{1 A} = 10 Ω$
As there is no current in $4 Ω$ , so there is no potential drop across it.
$∴$ The potential drop across the capacitor
$= 4 V - 2 V = 2 V$
The charge on the capacitor is
$Q = C V = (2 μ F) (2 V) = 4 μ C$

PHXII03:CURRENT ELECTRICITY

357483 In the circuit shown in the figure, the steady state voltage drop across the capacitor is $V_{C}$ . Then
supporting img

1

V_{C} = \frac{V R_{1}}{R_{2} + R_{3}}

2

V_{C} = \frac{V R_{2}}{R_{1} + R_{3}}

3

V_{C} = \frac{V R_{1}}{R_{1} + R_{2}}

4

V_{C} = \frac{V R_{2}}{R_{1} + R_{2}}

Explanation:

In the steady state no current will flow in the branch containing the capacitor.
{ $∴ R_{2}$ and are in parallel }
Hence $I = \frac{V}{R_{1} + R_{2}}$
However, the potential across the capacitor will be equal to the potential difference across the resistor $R_{2} .$
$\Rightarrow V_{C} = (\frac{R_{2}}{R_{1} + R_{2}}) V$

PHXII03:CURRENT ELECTRICITY

357484 A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ . The amount of charges on the capacitor will be
supporting img

1

4 μ C

2

16 μ C

3

0

4

8 μ C

Explanation:

No current is allowed to flow through $10 Ω$ resistance because of capacitor. The current drawn by $2 Ω$ resistance.
$I = \frac{E}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
Terminal potential difference of battery, $V = E - I r$
$= 2.5 - 1 \times 0.5 = 2 v o l t$
$∴$ Charge on capacitor plates,
$Q = C V = 4 \times 2 = 8 μ C$

PHXII03:CURRENT ELECTRICITY

357485 In the network shown below, the charge accumulated in the capacitor in steady state will be
supporting img

1

12 μ C

2

10.3 μ C

3

7.2 μ C

4

4.8 μ C

Explanation:

Steady state of capacitor $\to$ when there is no current flowing through it, the capacitor will behave as an open circuit.
The equivalent circuit diagram is given by, (As there is no current following through the capacitor and $6 Ω$ resistor) Current flowing through the circuit is given by,
supporting img
$I = \frac{V}{R} = \frac{3}{4 + 6} = \frac{3}{10} A$ .
Voltage across $6 Ω$ resistor and capacitor is same, as shown in the diagram
$V = \frac{3}{10} \times 6$
$V = 1.8 V$
The charge accumulated in capacitor in its steady state is, $Q = C V Q = 4 \times 10^{- 6} \times 1.8 C Q = 7.2 μ C$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357482 A network of resistances, cell and capacitor $C (= 2 μ F)$ is shown in figure. In steady state condition, the charge on $2 μ F$ capacitor is $Q$ , while $R$ is unknown resistance. Values of $Q$ and $R$ are respectively
supporting img

1

4 μ C

and

10 Ω

2

4 μ C

and

4 Ω

3

2 μ C

and

2 Ω

4

8 μ C

and

4 Ω

Explanation:

In the steady state, capacitor acts as a open circuit, hence no current will flow through $4 Ω$ .
$∴$ The currents distribution is shown in figure.
supporting img
The potential difference between $C$ and $D$ is
$V_{C} - V_{D} = (1 A) (2 Ω) = 2 V$
As $V_{D} = 0 V$ ( $D$ is earthed)
$∴ V_{C} = 2 V$
The potential drop across the resistance $R$
$= 10 V + 2 V - 2 V = 10 V$
$∴$ Resistance, $R = \frac{10 V}{1 A} = 10 Ω$
As there is no current in $4 Ω$ , so there is no potential drop across it.
$∴$ The potential drop across the capacitor
$= 4 V - 2 V = 2 V$
The charge on the capacitor is
$Q = C V = (2 μ F) (2 V) = 4 μ C$

PHXII03:CURRENT ELECTRICITY

357483 In the circuit shown in the figure, the steady state voltage drop across the capacitor is $V_{C}$ . Then
supporting img

1

V_{C} = \frac{V R_{1}}{R_{2} + R_{3}}

2

V_{C} = \frac{V R_{2}}{R_{1} + R_{3}}

3

V_{C} = \frac{V R_{1}}{R_{1} + R_{2}}

4

V_{C} = \frac{V R_{2}}{R_{1} + R_{2}}

Explanation:

In the steady state no current will flow in the branch containing the capacitor.
{ $∴ R_{2}$ and are in parallel }
Hence $I = \frac{V}{R_{1} + R_{2}}$
However, the potential across the capacitor will be equal to the potential difference across the resistor $R_{2} .$
$\Rightarrow V_{C} = (\frac{R_{2}}{R_{1} + R_{2}}) V$

PHXII03:CURRENT ELECTRICITY

357484 A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ . The amount of charges on the capacitor will be
supporting img

1

4 μ C

2

16 μ C

3

0

4

8 μ C

Explanation:

No current is allowed to flow through $10 Ω$ resistance because of capacitor. The current drawn by $2 Ω$ resistance.
$I = \frac{E}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
Terminal potential difference of battery, $V = E - I r$
$= 2.5 - 1 \times 0.5 = 2 v o l t$
$∴$ Charge on capacitor plates,
$Q = C V = 4 \times 2 = 8 μ C$

PHXII03:CURRENT ELECTRICITY

357485 In the network shown below, the charge accumulated in the capacitor in steady state will be
supporting img

1

12 μ C

2

10.3 μ C

3

7.2 μ C

4

4.8 μ C

Explanation:

Steady state of capacitor $\to$ when there is no current flowing through it, the capacitor will behave as an open circuit.
The equivalent circuit diagram is given by, (As there is no current following through the capacitor and $6 Ω$ resistor) Current flowing through the circuit is given by,
supporting img
$I = \frac{V}{R} = \frac{3}{4 + 6} = \frac{3}{10} A$ .
Voltage across $6 Ω$ resistor and capacitor is same, as shown in the diagram
$V = \frac{3}{10} \times 6$
$V = 1.8 V$
The charge accumulated in capacitor in its steady state is, $Q = C V Q = 4 \times 10^{- 6} \times 1.8 C Q = 7.2 μ C$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357482 A network of resistances, cell and capacitor $C (= 2 μ F)$ is shown in figure. In steady state condition, the charge on $2 μ F$ capacitor is $Q$ , while $R$ is unknown resistance. Values of $Q$ and $R$ are respectively
supporting img

1

4 μ C

and

10 Ω

2

4 μ C

and

4 Ω

3

2 μ C

and

2 Ω

4

8 μ C

and

4 Ω

Explanation:

In the steady state, capacitor acts as a open circuit, hence no current will flow through $4 Ω$ .
$∴$ The currents distribution is shown in figure.
supporting img
The potential difference between $C$ and $D$ is
$V_{C} - V_{D} = (1 A) (2 Ω) = 2 V$
As $V_{D} = 0 V$ ( $D$ is earthed)
$∴ V_{C} = 2 V$
The potential drop across the resistance $R$
$= 10 V + 2 V - 2 V = 10 V$
$∴$ Resistance, $R = \frac{10 V}{1 A} = 10 Ω$
As there is no current in $4 Ω$ , so there is no potential drop across it.
$∴$ The potential drop across the capacitor
$= 4 V - 2 V = 2 V$
The charge on the capacitor is
$Q = C V = (2 μ F) (2 V) = 4 μ C$

PHXII03:CURRENT ELECTRICITY

357483 In the circuit shown in the figure, the steady state voltage drop across the capacitor is $V_{C}$ . Then
supporting img

1

V_{C} = \frac{V R_{1}}{R_{2} + R_{3}}

2

V_{C} = \frac{V R_{2}}{R_{1} + R_{3}}

3

V_{C} = \frac{V R_{1}}{R_{1} + R_{2}}

4

V_{C} = \frac{V R_{2}}{R_{1} + R_{2}}

Explanation:

In the steady state no current will flow in the branch containing the capacitor.
{ $∴ R_{2}$ and are in parallel }
Hence $I = \frac{V}{R_{1} + R_{2}}$
However, the potential across the capacitor will be equal to the potential difference across the resistor $R_{2} .$
$\Rightarrow V_{C} = (\frac{R_{2}}{R_{1} + R_{2}}) V$

PHXII03:CURRENT ELECTRICITY

357484 A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ . The amount of charges on the capacitor will be
supporting img

1

4 μ C

2

16 μ C

3

0

4

8 μ C

Explanation:

No current is allowed to flow through $10 Ω$ resistance because of capacitor. The current drawn by $2 Ω$ resistance.
$I = \frac{E}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
Terminal potential difference of battery, $V = E - I r$
$= 2.5 - 1 \times 0.5 = 2 v o l t$
$∴$ Charge on capacitor plates,
$Q = C V = 4 \times 2 = 8 μ C$

PHXII03:CURRENT ELECTRICITY

357485 In the network shown below, the charge accumulated in the capacitor in steady state will be
supporting img

1

12 μ C

2

10.3 μ C

3

7.2 μ C

4

4.8 μ C

Explanation:

Steady state of capacitor $\to$ when there is no current flowing through it, the capacitor will behave as an open circuit.
The equivalent circuit diagram is given by, (As there is no current following through the capacitor and $6 Ω$ resistor) Current flowing through the circuit is given by,
supporting img
$I = \frac{V}{R} = \frac{3}{4 + 6} = \frac{3}{10} A$ .
Voltage across $6 Ω$ resistor and capacitor is same, as shown in the diagram
$V = \frac{3}{10} \times 6$
$V = 1.8 V$
The charge accumulated in capacitor in its steady state is, $Q = C V Q = 4 \times 10^{- 6} \times 1.8 C Q = 7.2 μ C$

JEE - 2023

PHXII03:CURRENT ELECTRICITY

357482 A network of resistances, cell and capacitor $C (= 2 μ F)$ is shown in figure. In steady state condition, the charge on $2 μ F$ capacitor is $Q$ , while $R$ is unknown resistance. Values of $Q$ and $R$ are respectively
supporting img

1

4 μ C

and

10 Ω

2

4 μ C

and

4 Ω

3

2 μ C

and

2 Ω

4

8 μ C

and

4 Ω

Explanation:

In the steady state, capacitor acts as a open circuit, hence no current will flow through $4 Ω$ .
$∴$ The currents distribution is shown in figure.
supporting img
The potential difference between $C$ and $D$ is
$V_{C} - V_{D} = (1 A) (2 Ω) = 2 V$
As $V_{D} = 0 V$ ( $D$ is earthed)
$∴ V_{C} = 2 V$
The potential drop across the resistance $R$
$= 10 V + 2 V - 2 V = 10 V$
$∴$ Resistance, $R = \frac{10 V}{1 A} = 10 Ω$
As there is no current in $4 Ω$ , so there is no potential drop across it.
$∴$ The potential drop across the capacitor
$= 4 V - 2 V = 2 V$
The charge on the capacitor is
$Q = C V = (2 μ F) (2 V) = 4 μ C$

PHXII03:CURRENT ELECTRICITY

357483 In the circuit shown in the figure, the steady state voltage drop across the capacitor is $V_{C}$ . Then
supporting img

1

V_{C} = \frac{V R_{1}}{R_{2} + R_{3}}

2

V_{C} = \frac{V R_{2}}{R_{1} + R_{3}}

3

V_{C} = \frac{V R_{1}}{R_{1} + R_{2}}

4

V_{C} = \frac{V R_{2}}{R_{1} + R_{2}}

Explanation:

In the steady state no current will flow in the branch containing the capacitor.
{ $∴ R_{2}$ and are in parallel }
Hence $I = \frac{V}{R_{1} + R_{2}}$
However, the potential across the capacitor will be equal to the potential difference across the resistor $R_{2} .$
$\Rightarrow V_{C} = (\frac{R_{2}}{R_{1} + R_{2}}) V$

PHXII03:CURRENT ELECTRICITY

357484 A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ . The amount of charges on the capacitor will be
supporting img

1

4 μ C

2

16 μ C

3

0

4

8 μ C

Explanation:

No current is allowed to flow through $10 Ω$ resistance because of capacitor. The current drawn by $2 Ω$ resistance.
$I = \frac{E}{R + r} = \frac{2.5}{2 + 0.5} = 1 A$
Terminal potential difference of battery, $V = E - I r$
$= 2.5 - 1 \times 0.5 = 2 v o l t$
$∴$ Charge on capacitor plates,
$Q = C V = 4 \times 2 = 8 μ C$

PHXII03:CURRENT ELECTRICITY

357485 In the network shown below, the charge accumulated in the capacitor in steady state will be
supporting img

1

12 μ C

2

10.3 μ C

3

7.2 μ C

4

4.8 μ C

Explanation:

Steady state of capacitor $\to$ when there is no current flowing through it, the capacitor will behave as an open circuit.
The equivalent circuit diagram is given by, (As there is no current following through the capacitor and $6 Ω$ resistor) Current flowing through the circuit is given by,
supporting img
$I = \frac{V}{R} = \frac{3}{4 + 6} = \frac{3}{10} A$ .
Voltage across $6 Ω$ resistor and capacitor is same, as shown in the diagram
$V = \frac{3}{10} \times 6$
$V = 1.8 V$
The charge accumulated in capacitor in its steady state is, $Q = C V Q = 4 \times 10^{- 6} \times 1.8 C Q = 7.2 μ C$

JEE - 2023