QBManager

PHXII03:CURRENT ELECTRICITY

357426 The potential difference $(V_{A} - V_{B})$ between the points $A$ and $B$ in the given figure is
supporting img

1

- 3 V

2

+ 3 V

3

+ 6 V

4

+ 9 V

Explanation:

$V_{A} - V_{B} = 1 \times 2 + 3 + 2 \times 2 = 9 V$

NEET - 2016

PHXII03:CURRENT ELECTRICITY

357427 In the circuit shown, current (in $A$ ) through the 50 $V$ and 30 $V$ batteries are respectively :
supporting img

1 2.5 and 3

2 3 and 2.5

3 3.5 and 2

4 4.5 and 1

Explanation:

Let $I_{1}$ and $I_{2}$ are currents through the circuit
supporting img
$K V L$ in loop $a b g h a$
$20 I_{1} = 50 \Rightarrow I_{1} = 2.5 A$
$K V L$ in loop $a b c d e f g h a$
$50 - 5 I_{2} - 30 - 5 I_{2} = 0 \Rightarrow I_{2} = 2 A$
$K V L$ in loop $c d e f c$
$30 = 10 (I_{2} + I_{3})$
$\Rightarrow I_{2} + I_{3} = 3 \Rightarrow I_{3} = 3 - 2 = 1 A$
Current through 50 $V$ battery is $= I_{1} + I_{2} = 4.5 A$
Current through 30 $V$ battery $= I_{3} = l A$

JEE - 2014

PHXII03:CURRENT ELECTRICITY

357428 A current of 6 $A$ enter one corner $P$ of an equilateral triangle $P Q R$ having 3 wires of resistances $2 Ω$ each and leaves by the corner $R$ . Then the current $I_{1} a n d I_{2}$ are
supporting img

1

2 A, 4 A

2

4 A, 2 A

3

1 A, 2 A

4

2 A, 3 A

Explanation:

Applying Kirchhoff's first law at the junction $P$ , we get
$6 = I_{1} + I_{2} (1)$
Applying Kirchhoff's secod law to the closed circuit PQRP, we get
$- 2 I - 2 I_{1} + 2 I_{2} = 0$
$o r, I_{2} = 2 I_{1} (2)$
Solve (1) and (2), we get
$I_{1} = 2 A, I_{2} = 4 A$

PHXII03:CURRENT ELECTRICITY

357429 Find out the value of current through $2 Ω$ resistance for the given circuit
supporting img

1

5 A

2

2 A

3

Z e r o

4

4 A

Explanation:

Since the current coming out from the positive terminal is equal to the current entering the negative terminal, therefore, current in the respective loops will remain confined in the loop itself. Current does not flow from one loop to the other loop.
$∴$ Current through $2 Ω$ resistor $= 0$

PHXII03:CURRENT ELECTRICITY

357426 The potential difference $(V_{A} - V_{B})$ between the points $A$ and $B$ in the given figure is
supporting img

1

- 3 V

2

+ 3 V

3

+ 6 V

4

+ 9 V

Explanation:

$V_{A} - V_{B} = 1 \times 2 + 3 + 2 \times 2 = 9 V$

NEET - 2016

PHXII03:CURRENT ELECTRICITY

357427 In the circuit shown, current (in $A$ ) through the 50 $V$ and 30 $V$ batteries are respectively :
supporting img

1 2.5 and 3

2 3 and 2.5

3 3.5 and 2

4 4.5 and 1

Explanation:

Let $I_{1}$ and $I_{2}$ are currents through the circuit
supporting img
$K V L$ in loop $a b g h a$
$20 I_{1} = 50 \Rightarrow I_{1} = 2.5 A$
$K V L$ in loop $a b c d e f g h a$
$50 - 5 I_{2} - 30 - 5 I_{2} = 0 \Rightarrow I_{2} = 2 A$
$K V L$ in loop $c d e f c$
$30 = 10 (I_{2} + I_{3})$
$\Rightarrow I_{2} + I_{3} = 3 \Rightarrow I_{3} = 3 - 2 = 1 A$
Current through 50 $V$ battery is $= I_{1} + I_{2} = 4.5 A$
Current through 30 $V$ battery $= I_{3} = l A$

JEE - 2014

PHXII03:CURRENT ELECTRICITY

357428 A current of 6 $A$ enter one corner $P$ of an equilateral triangle $P Q R$ having 3 wires of resistances $2 Ω$ each and leaves by the corner $R$ . Then the current $I_{1} a n d I_{2}$ are
supporting img

1

2 A, 4 A

2

4 A, 2 A

3

1 A, 2 A

4

2 A, 3 A

Explanation:

Applying Kirchhoff's first law at the junction $P$ , we get
$6 = I_{1} + I_{2} (1)$
Applying Kirchhoff's secod law to the closed circuit PQRP, we get
$- 2 I - 2 I_{1} + 2 I_{2} = 0$
$o r, I_{2} = 2 I_{1} (2)$
Solve (1) and (2), we get
$I_{1} = 2 A, I_{2} = 4 A$

PHXII03:CURRENT ELECTRICITY

357429 Find out the value of current through $2 Ω$ resistance for the given circuit
supporting img

1

5 A

2

2 A

3

Z e r o

4

4 A

Explanation:

Since the current coming out from the positive terminal is equal to the current entering the negative terminal, therefore, current in the respective loops will remain confined in the loop itself. Current does not flow from one loop to the other loop.
$∴$ Current through $2 Ω$ resistor $= 0$

PHXII03:CURRENT ELECTRICITY

357426 The potential difference $(V_{A} - V_{B})$ between the points $A$ and $B$ in the given figure is
supporting img

1

- 3 V

2

+ 3 V

3

+ 6 V

4

+ 9 V

Explanation:

$V_{A} - V_{B} = 1 \times 2 + 3 + 2 \times 2 = 9 V$

NEET - 2016

PHXII03:CURRENT ELECTRICITY

357427 In the circuit shown, current (in $A$ ) through the 50 $V$ and 30 $V$ batteries are respectively :
supporting img

1 2.5 and 3

2 3 and 2.5

3 3.5 and 2

4 4.5 and 1

Explanation:

Let $I_{1}$ and $I_{2}$ are currents through the circuit
supporting img
$K V L$ in loop $a b g h a$
$20 I_{1} = 50 \Rightarrow I_{1} = 2.5 A$
$K V L$ in loop $a b c d e f g h a$
$50 - 5 I_{2} - 30 - 5 I_{2} = 0 \Rightarrow I_{2} = 2 A$
$K V L$ in loop $c d e f c$
$30 = 10 (I_{2} + I_{3})$
$\Rightarrow I_{2} + I_{3} = 3 \Rightarrow I_{3} = 3 - 2 = 1 A$
Current through 50 $V$ battery is $= I_{1} + I_{2} = 4.5 A$
Current through 30 $V$ battery $= I_{3} = l A$

JEE - 2014

PHXII03:CURRENT ELECTRICITY

357428 A current of 6 $A$ enter one corner $P$ of an equilateral triangle $P Q R$ having 3 wires of resistances $2 Ω$ each and leaves by the corner $R$ . Then the current $I_{1} a n d I_{2}$ are
supporting img

1

2 A, 4 A

2

4 A, 2 A

3

1 A, 2 A

4

2 A, 3 A

Explanation:

Applying Kirchhoff's first law at the junction $P$ , we get
$6 = I_{1} + I_{2} (1)$
Applying Kirchhoff's secod law to the closed circuit PQRP, we get
$- 2 I - 2 I_{1} + 2 I_{2} = 0$
$o r, I_{2} = 2 I_{1} (2)$
Solve (1) and (2), we get
$I_{1} = 2 A, I_{2} = 4 A$

PHXII03:CURRENT ELECTRICITY

357429 Find out the value of current through $2 Ω$ resistance for the given circuit
supporting img

1

5 A

2

2 A

3

Z e r o

4

4 A

Explanation:

Since the current coming out from the positive terminal is equal to the current entering the negative terminal, therefore, current in the respective loops will remain confined in the loop itself. Current does not flow from one loop to the other loop.
$∴$ Current through $2 Ω$ resistor $= 0$

PHXII03:CURRENT ELECTRICITY

357426 The potential difference $(V_{A} - V_{B})$ between the points $A$ and $B$ in the given figure is
supporting img

1

- 3 V

2

+ 3 V

3

+ 6 V

4

+ 9 V

Explanation:

$V_{A} - V_{B} = 1 \times 2 + 3 + 2 \times 2 = 9 V$

NEET - 2016

PHXII03:CURRENT ELECTRICITY

357427 In the circuit shown, current (in $A$ ) through the 50 $V$ and 30 $V$ batteries are respectively :
supporting img

1 2.5 and 3

2 3 and 2.5

3 3.5 and 2

4 4.5 and 1

Explanation:

Let $I_{1}$ and $I_{2}$ are currents through the circuit
supporting img
$K V L$ in loop $a b g h a$
$20 I_{1} = 50 \Rightarrow I_{1} = 2.5 A$
$K V L$ in loop $a b c d e f g h a$
$50 - 5 I_{2} - 30 - 5 I_{2} = 0 \Rightarrow I_{2} = 2 A$
$K V L$ in loop $c d e f c$
$30 = 10 (I_{2} + I_{3})$
$\Rightarrow I_{2} + I_{3} = 3 \Rightarrow I_{3} = 3 - 2 = 1 A$
Current through 50 $V$ battery is $= I_{1} + I_{2} = 4.5 A$
Current through 30 $V$ battery $= I_{3} = l A$

JEE - 2014

PHXII03:CURRENT ELECTRICITY

357428 A current of 6 $A$ enter one corner $P$ of an equilateral triangle $P Q R$ having 3 wires of resistances $2 Ω$ each and leaves by the corner $R$ . Then the current $I_{1} a n d I_{2}$ are
supporting img

1

2 A, 4 A

2

4 A, 2 A

3

1 A, 2 A

4

2 A, 3 A

Explanation:

Applying Kirchhoff's first law at the junction $P$ , we get
$6 = I_{1} + I_{2} (1)$
Applying Kirchhoff's secod law to the closed circuit PQRP, we get
$- 2 I - 2 I_{1} + 2 I_{2} = 0$
$o r, I_{2} = 2 I_{1} (2)$
Solve (1) and (2), we get
$I_{1} = 2 A, I_{2} = 4 A$

PHXII03:CURRENT ELECTRICITY

357429 Find out the value of current through $2 Ω$ resistance for the given circuit
supporting img

1

5 A

2

2 A

3

Z e r o

4

4 A

Explanation:

Since the current coming out from the positive terminal is equal to the current entering the negative terminal, therefore, current in the respective loops will remain confined in the loop itself. Current does not flow from one loop to the other loop.
$∴$ Current through $2 Ω$ resistor $= 0$