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PHXII03:CURRENT ELECTRICITY

357396 A part of a circuit is shown in figure. $V_{B} - V_{C}$ is equal to $12 V . I_{1}$ is equal to
supporting img

1

1 A

2

2 A

3

3 A

4

4 A

Explanation:

If $V_{B} - V_{C} = 12$ then potential drop across $2 Ω$ resistor present between $B$ and $C$ is $2 V$ so, current flow from $B$ to $C$ .
$= \frac{2}{2} = 1 A$ . By junction rule $I_{1} = 2 + 1 = 3 A$ .

PHXII03:CURRENT ELECTRICITY

357397 The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is
supporting img

1 11

2 9

3 15

4 5

Explanation:

At junction $C : i_{1} + i_{2} = i$ , where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$
supporting img
Now $V_{A} - i_{1} + 1 + 1 = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V = \frac{- 22}{x} V$
$\Rightarrow x = 9$

PHXII03:CURRENT ELECTRICITY

357398 In the circuit shown, the current in the $2 Ω$ resistor is
supporting img

1

1 A

2

4 A

3

2 A

4

8 A

Explanation:

supporting img
Applying KVL to loop APQBA,
$2 I_{1} + (I_{1} - I_{2}) + 3 I_{1} = 10$
$6 I_{1} - I_{2} = 10$
$∴ I_{2} = 6 I_{1} - 10$ $(1)$
Considering loop $C D Q P C$ ,
$3 I_{2} - (I_{1} - I_{2}) = 6$
$4 I_{2} - I_{1} = 6$
Using equation (1),
$4 (6 I_{1} - 10) - I_{1} = 6$
$23 I_{1} - 40 = 6$
$∴ I_{1} = \frac{46}{23} A$
$∴ I_{1} = 2 A$

PHXII03:CURRENT ELECTRICITY

357399 Resistors of $6 Ω$ and $12 Ω$ are connected in parallel. This combination is connected in series with a $10 V$ battery and $6 Ω$ resistor. What is the potential difference between the terminals of the $12 Ω$ resistor?

1

4 V

2

16 V

3

2 V

4

8 V

Explanation:

The circuit is shown as below
supporting img
Here, $6 Ω$ and $12 Ω$ are in parallel.
So, $R = \frac{6 \times 12}{6 + 12} = \frac{6 \times 12}{18} = 4 Ω$
Total resistance,
$R_{e q} = 6 + 4 = 10 Ω$
$∴$ Current, $i = \frac{V}{R} = \frac{10}{10} = 1 A$
The current in $12 Ω$ resistor is
$i_{2} = i (\frac{R_{1}}{R_{1} + R_{2}}) = 1 \times (\frac{6}{6 + 12})$
$\Rightarrow i_{2} = \frac{1}{3} A$
The potential difference in $12 Ω$ resistor,
$V = i_{2} R = \frac{1}{3} \times 12 = 4 V$
So, correct option is (1)

AIIMS - 2012

PHXII03:CURRENT ELECTRICITY

357396 A part of a circuit is shown in figure. $V_{B} - V_{C}$ is equal to $12 V . I_{1}$ is equal to
supporting img

1

1 A

2

2 A

3

3 A

4

4 A

Explanation:

If $V_{B} - V_{C} = 12$ then potential drop across $2 Ω$ resistor present between $B$ and $C$ is $2 V$ so, current flow from $B$ to $C$ .
$= \frac{2}{2} = 1 A$ . By junction rule $I_{1} = 2 + 1 = 3 A$ .

PHXII03:CURRENT ELECTRICITY

357397 The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is
supporting img

1 11

2 9

3 15

4 5

Explanation:

At junction $C : i_{1} + i_{2} = i$ , where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$
supporting img
Now $V_{A} - i_{1} + 1 + 1 = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V = \frac{- 22}{x} V$
$\Rightarrow x = 9$

PHXII03:CURRENT ELECTRICITY

357398 In the circuit shown, the current in the $2 Ω$ resistor is
supporting img

1

1 A

2

4 A

3

2 A

4

8 A

Explanation:

supporting img
Applying KVL to loop APQBA,
$2 I_{1} + (I_{1} - I_{2}) + 3 I_{1} = 10$
$6 I_{1} - I_{2} = 10$
$∴ I_{2} = 6 I_{1} - 10$ $(1)$
Considering loop $C D Q P C$ ,
$3 I_{2} - (I_{1} - I_{2}) = 6$
$4 I_{2} - I_{1} = 6$
Using equation (1),
$4 (6 I_{1} - 10) - I_{1} = 6$
$23 I_{1} - 40 = 6$
$∴ I_{1} = \frac{46}{23} A$
$∴ I_{1} = 2 A$

PHXII03:CURRENT ELECTRICITY

357399 Resistors of $6 Ω$ and $12 Ω$ are connected in parallel. This combination is connected in series with a $10 V$ battery and $6 Ω$ resistor. What is the potential difference between the terminals of the $12 Ω$ resistor?

1

4 V

2

16 V

3

2 V

4

8 V

Explanation:

The circuit is shown as below
supporting img
Here, $6 Ω$ and $12 Ω$ are in parallel.
So, $R = \frac{6 \times 12}{6 + 12} = \frac{6 \times 12}{18} = 4 Ω$
Total resistance,
$R_{e q} = 6 + 4 = 10 Ω$
$∴$ Current, $i = \frac{V}{R} = \frac{10}{10} = 1 A$
The current in $12 Ω$ resistor is
$i_{2} = i (\frac{R_{1}}{R_{1} + R_{2}}) = 1 \times (\frac{6}{6 + 12})$
$\Rightarrow i_{2} = \frac{1}{3} A$
The potential difference in $12 Ω$ resistor,
$V = i_{2} R = \frac{1}{3} \times 12 = 4 V$
So, correct option is (1)

AIIMS - 2012

PHXII03:CURRENT ELECTRICITY

357396 A part of a circuit is shown in figure. $V_{B} - V_{C}$ is equal to $12 V . I_{1}$ is equal to
supporting img

1

1 A

2

2 A

3

3 A

4

4 A

Explanation:

If $V_{B} - V_{C} = 12$ then potential drop across $2 Ω$ resistor present between $B$ and $C$ is $2 V$ so, current flow from $B$ to $C$ .
$= \frac{2}{2} = 1 A$ . By junction rule $I_{1} = 2 + 1 = 3 A$ .

PHXII03:CURRENT ELECTRICITY

357397 The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is
supporting img

1 11

2 9

3 15

4 5

Explanation:

At junction $C : i_{1} + i_{2} = i$ , where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$
supporting img
Now $V_{A} - i_{1} + 1 + 1 = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V = \frac{- 22}{x} V$
$\Rightarrow x = 9$

PHXII03:CURRENT ELECTRICITY

357398 In the circuit shown, the current in the $2 Ω$ resistor is
supporting img

1

1 A

2

4 A

3

2 A

4

8 A

Explanation:

supporting img
Applying KVL to loop APQBA,
$2 I_{1} + (I_{1} - I_{2}) + 3 I_{1} = 10$
$6 I_{1} - I_{2} = 10$
$∴ I_{2} = 6 I_{1} - 10$ $(1)$
Considering loop $C D Q P C$ ,
$3 I_{2} - (I_{1} - I_{2}) = 6$
$4 I_{2} - I_{1} = 6$
Using equation (1),
$4 (6 I_{1} - 10) - I_{1} = 6$
$23 I_{1} - 40 = 6$
$∴ I_{1} = \frac{46}{23} A$
$∴ I_{1} = 2 A$

PHXII03:CURRENT ELECTRICITY

357399 Resistors of $6 Ω$ and $12 Ω$ are connected in parallel. This combination is connected in series with a $10 V$ battery and $6 Ω$ resistor. What is the potential difference between the terminals of the $12 Ω$ resistor?

1

4 V

2

16 V

3

2 V

4

8 V

Explanation:

The circuit is shown as below
supporting img
Here, $6 Ω$ and $12 Ω$ are in parallel.
So, $R = \frac{6 \times 12}{6 + 12} = \frac{6 \times 12}{18} = 4 Ω$
Total resistance,
$R_{e q} = 6 + 4 = 10 Ω$
$∴$ Current, $i = \frac{V}{R} = \frac{10}{10} = 1 A$
The current in $12 Ω$ resistor is
$i_{2} = i (\frac{R_{1}}{R_{1} + R_{2}}) = 1 \times (\frac{6}{6 + 12})$
$\Rightarrow i_{2} = \frac{1}{3} A$
The potential difference in $12 Ω$ resistor,
$V = i_{2} R = \frac{1}{3} \times 12 = 4 V$
So, correct option is (1)

AIIMS - 2012

PHXII03:CURRENT ELECTRICITY

357396 A part of a circuit is shown in figure. $V_{B} - V_{C}$ is equal to $12 V . I_{1}$ is equal to
supporting img

1

1 A

2

2 A

3

3 A

4

4 A

Explanation:

If $V_{B} - V_{C} = 12$ then potential drop across $2 Ω$ resistor present between $B$ and $C$ is $2 V$ so, current flow from $B$ to $C$ .
$= \frac{2}{2} = 1 A$ . By junction rule $I_{1} = 2 + 1 = 3 A$ .

PHXII03:CURRENT ELECTRICITY

357397 The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is
supporting img

1 11

2 9

3 15

4 5

Explanation:

At junction $C : i_{1} + i_{2} = i$ , where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$
supporting img
Now $V_{A} - i_{1} + 1 + 1 = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V = \frac{- 22}{x} V$
$\Rightarrow x = 9$

PHXII03:CURRENT ELECTRICITY

357398 In the circuit shown, the current in the $2 Ω$ resistor is
supporting img

1

1 A

2

4 A

3

2 A

4

8 A

Explanation:

supporting img
Applying KVL to loop APQBA,
$2 I_{1} + (I_{1} - I_{2}) + 3 I_{1} = 10$
$6 I_{1} - I_{2} = 10$
$∴ I_{2} = 6 I_{1} - 10$ $(1)$
Considering loop $C D Q P C$ ,
$3 I_{2} - (I_{1} - I_{2}) = 6$
$4 I_{2} - I_{1} = 6$
Using equation (1),
$4 (6 I_{1} - 10) - I_{1} = 6$
$23 I_{1} - 40 = 6$
$∴ I_{1} = \frac{46}{23} A$
$∴ I_{1} = 2 A$

PHXII03:CURRENT ELECTRICITY

357399 Resistors of $6 Ω$ and $12 Ω$ are connected in parallel. This combination is connected in series with a $10 V$ battery and $6 Ω$ resistor. What is the potential difference between the terminals of the $12 Ω$ resistor?

1

4 V

2

16 V

3

2 V

4

8 V

Explanation:

The circuit is shown as below
supporting img
Here, $6 Ω$ and $12 Ω$ are in parallel.
So, $R = \frac{6 \times 12}{6 + 12} = \frac{6 \times 12}{18} = 4 Ω$
Total resistance,
$R_{e q} = 6 + 4 = 10 Ω$
$∴$ Current, $i = \frac{V}{R} = \frac{10}{10} = 1 A$
The current in $12 Ω$ resistor is
$i_{2} = i (\frac{R_{1}}{R_{1} + R_{2}}) = 1 \times (\frac{6}{6 + 12})$
$\Rightarrow i_{2} = \frac{1}{3} A$
The potential difference in $12 Ω$ resistor,
$V = i_{2} R = \frac{1}{3} \times 12 = 4 V$
So, correct option is (1)

AIIMS - 2012

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