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PHXII03:CURRENT ELECTRICITY

357486 In the circuit shown in figure, with steady current, the potential drop across the capacitor must be
supporting img

1

\frac{2 V}{3}

2

\frac{V}{2}

3

V

4

\frac{V}{3}

Explanation:

Applying Kirchhoff’s law in $B C D E F A B$ we get, $I = \frac{V}{3 R}$
Let $A$ be at 0 $V$ . Then potential at $G$ is $V$ .
Applying Krichhoff’s law for $A F E D$ , we get
$0 + V + I R = V_{D}$
$\Rightarrow 0 + V + \frac{V}{3 R} \times R = V_{D} \Rightarrow V_{D} = \frac{4 V}{3}$
$V_{D} - V_{G} = \frac{V}{3}$
supporting img

PHXII03:CURRENT ELECTRICITY

357487 In the circuit shown switch is closed at time $t = 0$ . Let $I_{1}$ and $I_{2}$ be the currents in both branches at $t = 0$ , then ratio of $\frac{I_{1}}{I_{2}}$ .
supporting img

1

\frac{I_{1}}{I_{2}} = 2 : 1

2

= 1 : 2

3

\frac{I_{1}}{I_{2}} = 1 : 1

4

\frac{I_{1}}{I_{2}} = \frac{2 R C}{R C}

Explanation:

At $t = 0$ capacitors can be replaced with conducting wires and hence current in
both branches $I = \frac{V}{R}$ .
$\Rightarrow \frac{I_{1}}{I_{2}} = 1 : 1$

PHXII03:CURRENT ELECTRICITY

357488 Calculate the charge on capacitor $A$ in the circuit shown in figure in steady state.
supporting img

1

16 μ C

2

36 μ C

3

54 μ C

4

72 μ C

Explanation:

The current distribution is shown in figure. As we know that in steady state no current will flow in the two branches consisting of capacitors.
From figure the currents in the two loops are given as
$i_{1} = \frac{24}{4 + 2} = 4 A and i_{2} = \frac{18}{3 + 6} = 2 A$
In the figure we consider a potential $x$ at the right plate of the $2 m F$ capacitor and zero at the left plate of this capacitor and find the potentials at the plates of other capacitor as shown in the figure.
supporting img
Now writing nodal equation for unknown potential $x$ , we have
$\begin{array}{ll} 2 x + 3 (x - 12 - 18) = 0 \\ \Rightarrow & 5 x = 90 \\ \Rightarrow & x = 18 V \end{array}$
Charge on $2 μ F$ capacitor is given as $q_{2 μ F} = 2 \times 18 = 36 μ C$
Charge on $3 μ F$ capacitor is given as $q_{3 μ F} = 3 \times (18 - 6) = 36 μ C$

PHXII03:CURRENT ELECTRICITY

357489 In the given figure each plate of capacitance $C$ has value of charge -
supporting img

1

C E

2

\frac{C E R_{1}}{R_{2} - r}

3

\frac{C E R_{2}}{R_{2} + r}

4

\frac{C E R_{1}}{R_{1} - r}

Explanation:

$I = \frac{E}{r + R_{2}} P . D across R_{2} = I R_{2} = \frac{E R_{2}}{r + R_{2}}$
$Q = \frac{C E R_{2}}{r + R_{2}}$

PHXII03:CURRENT ELECTRICITY

357486 In the circuit shown in figure, with steady current, the potential drop across the capacitor must be
supporting img

1

\frac{2 V}{3}

2

\frac{V}{2}

3

V

4

\frac{V}{3}

Explanation:

Applying Kirchhoff’s law in $B C D E F A B$ we get, $I = \frac{V}{3 R}$
Let $A$ be at 0 $V$ . Then potential at $G$ is $V$ .
Applying Krichhoff’s law for $A F E D$ , we get
$0 + V + I R = V_{D}$
$\Rightarrow 0 + V + \frac{V}{3 R} \times R = V_{D} \Rightarrow V_{D} = \frac{4 V}{3}$
$V_{D} - V_{G} = \frac{V}{3}$
supporting img

PHXII03:CURRENT ELECTRICITY

357487 In the circuit shown switch is closed at time $t = 0$ . Let $I_{1}$ and $I_{2}$ be the currents in both branches at $t = 0$ , then ratio of $\frac{I_{1}}{I_{2}}$ .
supporting img

1

\frac{I_{1}}{I_{2}} = 2 : 1

2

= 1 : 2

3

\frac{I_{1}}{I_{2}} = 1 : 1

4

\frac{I_{1}}{I_{2}} = \frac{2 R C}{R C}

Explanation:

At $t = 0$ capacitors can be replaced with conducting wires and hence current in
both branches $I = \frac{V}{R}$ .
$\Rightarrow \frac{I_{1}}{I_{2}} = 1 : 1$

PHXII03:CURRENT ELECTRICITY

357488 Calculate the charge on capacitor $A$ in the circuit shown in figure in steady state.
supporting img

1

16 μ C

2

36 μ C

3

54 μ C

4

72 μ C

Explanation:

The current distribution is shown in figure. As we know that in steady state no current will flow in the two branches consisting of capacitors.
From figure the currents in the two loops are given as
$i_{1} = \frac{24}{4 + 2} = 4 A and i_{2} = \frac{18}{3 + 6} = 2 A$
In the figure we consider a potential $x$ at the right plate of the $2 m F$ capacitor and zero at the left plate of this capacitor and find the potentials at the plates of other capacitor as shown in the figure.
supporting img
Now writing nodal equation for unknown potential $x$ , we have
$\begin{array}{ll} 2 x + 3 (x - 12 - 18) = 0 \\ \Rightarrow & 5 x = 90 \\ \Rightarrow & x = 18 V \end{array}$
Charge on $2 μ F$ capacitor is given as $q_{2 μ F} = 2 \times 18 = 36 μ C$
Charge on $3 μ F$ capacitor is given as $q_{3 μ F} = 3 \times (18 - 6) = 36 μ C$

PHXII03:CURRENT ELECTRICITY

357489 In the given figure each plate of capacitance $C$ has value of charge -
supporting img

1

C E

2

\frac{C E R_{1}}{R_{2} - r}

3

\frac{C E R_{2}}{R_{2} + r}

4

\frac{C E R_{1}}{R_{1} - r}

Explanation:

$I = \frac{E}{r + R_{2}} P . D across R_{2} = I R_{2} = \frac{E R_{2}}{r + R_{2}}$
$Q = \frac{C E R_{2}}{r + R_{2}}$

PHXII03:CURRENT ELECTRICITY

357486 In the circuit shown in figure, with steady current, the potential drop across the capacitor must be
supporting img

1

\frac{2 V}{3}

2

\frac{V}{2}

3

V

4

\frac{V}{3}

Explanation:

Applying Kirchhoff’s law in $B C D E F A B$ we get, $I = \frac{V}{3 R}$
Let $A$ be at 0 $V$ . Then potential at $G$ is $V$ .
Applying Krichhoff’s law for $A F E D$ , we get
$0 + V + I R = V_{D}$
$\Rightarrow 0 + V + \frac{V}{3 R} \times R = V_{D} \Rightarrow V_{D} = \frac{4 V}{3}$
$V_{D} - V_{G} = \frac{V}{3}$
supporting img

PHXII03:CURRENT ELECTRICITY

357487 In the circuit shown switch is closed at time $t = 0$ . Let $I_{1}$ and $I_{2}$ be the currents in both branches at $t = 0$ , then ratio of $\frac{I_{1}}{I_{2}}$ .
supporting img

1

\frac{I_{1}}{I_{2}} = 2 : 1

2

= 1 : 2

3

\frac{I_{1}}{I_{2}} = 1 : 1

4

\frac{I_{1}}{I_{2}} = \frac{2 R C}{R C}

Explanation:

At $t = 0$ capacitors can be replaced with conducting wires and hence current in
both branches $I = \frac{V}{R}$ .
$\Rightarrow \frac{I_{1}}{I_{2}} = 1 : 1$

PHXII03:CURRENT ELECTRICITY

357488 Calculate the charge on capacitor $A$ in the circuit shown in figure in steady state.
supporting img

1

16 μ C

2

36 μ C

3

54 μ C

4

72 μ C

Explanation:

The current distribution is shown in figure. As we know that in steady state no current will flow in the two branches consisting of capacitors.
From figure the currents in the two loops are given as
$i_{1} = \frac{24}{4 + 2} = 4 A and i_{2} = \frac{18}{3 + 6} = 2 A$
In the figure we consider a potential $x$ at the right plate of the $2 m F$ capacitor and zero at the left plate of this capacitor and find the potentials at the plates of other capacitor as shown in the figure.
supporting img
Now writing nodal equation for unknown potential $x$ , we have
$\begin{array}{ll} 2 x + 3 (x - 12 - 18) = 0 \\ \Rightarrow & 5 x = 90 \\ \Rightarrow & x = 18 V \end{array}$
Charge on $2 μ F$ capacitor is given as $q_{2 μ F} = 2 \times 18 = 36 μ C$
Charge on $3 μ F$ capacitor is given as $q_{3 μ F} = 3 \times (18 - 6) = 36 μ C$

PHXII03:CURRENT ELECTRICITY

357489 In the given figure each plate of capacitance $C$ has value of charge -
supporting img

1

C E

2

\frac{C E R_{1}}{R_{2} - r}

3

\frac{C E R_{2}}{R_{2} + r}

4

\frac{C E R_{1}}{R_{1} - r}

Explanation:

$I = \frac{E}{r + R_{2}} P . D across R_{2} = I R_{2} = \frac{E R_{2}}{r + R_{2}}$
$Q = \frac{C E R_{2}}{r + R_{2}}$

PHXII03:CURRENT ELECTRICITY

357486 In the circuit shown in figure, with steady current, the potential drop across the capacitor must be
supporting img

1

\frac{2 V}{3}

2

\frac{V}{2}

3

V

4

\frac{V}{3}

Explanation:

Applying Kirchhoff’s law in $B C D E F A B$ we get, $I = \frac{V}{3 R}$
Let $A$ be at 0 $V$ . Then potential at $G$ is $V$ .
Applying Krichhoff’s law for $A F E D$ , we get
$0 + V + I R = V_{D}$
$\Rightarrow 0 + V + \frac{V}{3 R} \times R = V_{D} \Rightarrow V_{D} = \frac{4 V}{3}$
$V_{D} - V_{G} = \frac{V}{3}$
supporting img

PHXII03:CURRENT ELECTRICITY

357487 In the circuit shown switch is closed at time $t = 0$ . Let $I_{1}$ and $I_{2}$ be the currents in both branches at $t = 0$ , then ratio of $\frac{I_{1}}{I_{2}}$ .
supporting img

1

\frac{I_{1}}{I_{2}} = 2 : 1

2

= 1 : 2

3

\frac{I_{1}}{I_{2}} = 1 : 1

4

\frac{I_{1}}{I_{2}} = \frac{2 R C}{R C}

Explanation:

At $t = 0$ capacitors can be replaced with conducting wires and hence current in
both branches $I = \frac{V}{R}$ .
$\Rightarrow \frac{I_{1}}{I_{2}} = 1 : 1$

PHXII03:CURRENT ELECTRICITY

357488 Calculate the charge on capacitor $A$ in the circuit shown in figure in steady state.
supporting img

1

16 μ C

2

36 μ C

3

54 μ C

4

72 μ C

Explanation:

The current distribution is shown in figure. As we know that in steady state no current will flow in the two branches consisting of capacitors.
From figure the currents in the two loops are given as
$i_{1} = \frac{24}{4 + 2} = 4 A and i_{2} = \frac{18}{3 + 6} = 2 A$
In the figure we consider a potential $x$ at the right plate of the $2 m F$ capacitor and zero at the left plate of this capacitor and find the potentials at the plates of other capacitor as shown in the figure.
supporting img
Now writing nodal equation for unknown potential $x$ , we have
$\begin{array}{ll} 2 x + 3 (x - 12 - 18) = 0 \\ \Rightarrow & 5 x = 90 \\ \Rightarrow & x = 18 V \end{array}$
Charge on $2 μ F$ capacitor is given as $q_{2 μ F} = 2 \times 18 = 36 μ C$
Charge on $3 μ F$ capacitor is given as $q_{3 μ F} = 3 \times (18 - 6) = 36 μ C$

PHXII03:CURRENT ELECTRICITY

357489 In the given figure each plate of capacitance $C$ has value of charge -
supporting img

1

C E

2

\frac{C E R_{1}}{R_{2} - r}

3

\frac{C E R_{2}}{R_{2} + r}

4

\frac{C E R_{1}}{R_{1} - r}

Explanation:

$I = \frac{E}{r + R_{2}} P . D across R_{2} = I R_{2} = \frac{E R_{2}}{r + R_{2}}$
$Q = \frac{C E R_{2}}{r + R_{2}}$

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