QBManager

Current Electricity

151904 An electric motor operates on a $50 V$ supply and a current of $12 A$ . If the efficiency of the motor is $30 %$ , what is the resistance of the winding of the motor?

1

6 Ω

2

4 Ω

3

2.9 Ω

4

3.1 Ω

Explanation:

C Given,
$V = 50 Volt$
$I = 12 A$
Efficiency $(η) = 30 %$
Power Input $(P_{I}) = 50 \times 12$
$= 600 W$
Power output $(P_{o}) = \frac{30}{100} \times 600 = 180 W$
Net power, $P = (P_{I} - P_{o})$
$= 600 - 180 = 420 W$
We know that,
Power, $P = I^{2} R$
$420 = (12)^{2} \times R$
$R = \frac{420}{144} = 2.9 Ω$

Manipal UGET-2018

Current Electricity

151905 A uniform wire of resistance $9 Ω$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC$ . A cell of emf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1

1 V

2

2 V

3

3 V

4

0.5 V

Explanation:

A
$∵ 3 Ω$ and $3 Ω$ are series

In the circuit,
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = 2 Ω$
$R_{eq} = 2 Ω$
Now,
$V = IR$
$2 = I \times 2$
$I = 1 A$
$∴ i_{1} = 1 \times (\frac{3}{3 + 6})$
$i_{1} = \frac{1}{3} A$
Potential difference between $A$ and $B$ ,
$V_{A} - V_{B} = \frac{1}{3} \times R$
$V_{A} - V_{B} = \frac{1}{3} \times 3$
$V_{A} - V_{B} = 1 V$

Manipal UGET-2018

Current Electricity

151907 A circular coil of single turn has a resistance of $20 Ω$ . Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?

1

5 Ω

2

10 Ω

3

20 Ω

4

40 Ω

Explanation:

A Given, Resistance of circular coil of single turn $= 20 Ω$
If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel.

$∴$ Equivalent Resistance,
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10}$
$∴ R_{eq} = 5 Ω$

NDA (I) 2017

Current Electricity

151908 On stretching a wire its length is increased by $0.2 %$ , its resistance will

1 increase by

0.1 %

2 decrease by

0.1 %

3 increase by

0.2 %

4 increase by

0.4 %

Explanation:

D $L_{2} = L_{1} + 0.2 %$ of $L_{1}$
$L_{2} = 1.002 L_{1}$
$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{V}$
Where $A$ is area, and $V$ is volume
$R \propto L^{2}$
Current Electricity
$\frac{R_{2}}{R_{1}} = {(\frac{L_{2}}{L_{1}})}^{2}$
$\frac{R_{2}}{R_{1}} = {(\frac{1.002 L_{1}}{L_{1}})}^{2} = (1.002)^{2}$
$R_{2} = 1.004 R_{1}$
$R_{2}$ is increased by $0.4 %$ of $R_{1}$

AMU-2017

Current Electricity

151910 Six wires, each of resistance $r$ , are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

1

r

2

2 r

3

\frac{r}{3}

4

\frac{r}{2}

Explanation:

D
Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit.
$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r}$
$R = r$
It will in parallel with outer resistance
$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$
$R_{eq} = \frac{r}{2}$

WB JEE 2017

Current Electricity

151904 An electric motor operates on a $50 V$ supply and a current of $12 A$ . If the efficiency of the motor is $30 %$ , what is the resistance of the winding of the motor?

1

6 Ω

2

4 Ω

3

2.9 Ω

4

3.1 Ω

Explanation:

C Given,
$V = 50 Volt$
$I = 12 A$
Efficiency $(η) = 30 %$
Power Input $(P_{I}) = 50 \times 12$
$= 600 W$
Power output $(P_{o}) = \frac{30}{100} \times 600 = 180 W$
Net power, $P = (P_{I} - P_{o})$
$= 600 - 180 = 420 W$
We know that,
Power, $P = I^{2} R$
$420 = (12)^{2} \times R$
$R = \frac{420}{144} = 2.9 Ω$

Manipal UGET-2018

Current Electricity

151905 A uniform wire of resistance $9 Ω$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC$ . A cell of emf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1

1 V

2

2 V

3

3 V

4

0.5 V

Explanation:

A
$∵ 3 Ω$ and $3 Ω$ are series

In the circuit,
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = 2 Ω$
$R_{eq} = 2 Ω$
Now,
$V = IR$
$2 = I \times 2$
$I = 1 A$
$∴ i_{1} = 1 \times (\frac{3}{3 + 6})$
$i_{1} = \frac{1}{3} A$
Potential difference between $A$ and $B$ ,
$V_{A} - V_{B} = \frac{1}{3} \times R$
$V_{A} - V_{B} = \frac{1}{3} \times 3$
$V_{A} - V_{B} = 1 V$

Manipal UGET-2018

Current Electricity

151907 A circular coil of single turn has a resistance of $20 Ω$ . Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?

1

5 Ω

2

10 Ω

3

20 Ω

4

40 Ω

Explanation:

A Given, Resistance of circular coil of single turn $= 20 Ω$
If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel.

$∴$ Equivalent Resistance,
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10}$
$∴ R_{eq} = 5 Ω$

NDA (I) 2017

Current Electricity

151908 On stretching a wire its length is increased by $0.2 %$ , its resistance will

1 increase by

0.1 %

2 decrease by

0.1 %

3 increase by

0.2 %

4 increase by

0.4 %

Explanation:

D $L_{2} = L_{1} + 0.2 %$ of $L_{1}$
$L_{2} = 1.002 L_{1}$
$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{V}$
Where $A$ is area, and $V$ is volume
$R \propto L^{2}$
Current Electricity
$\frac{R_{2}}{R_{1}} = {(\frac{L_{2}}{L_{1}})}^{2}$
$\frac{R_{2}}{R_{1}} = {(\frac{1.002 L_{1}}{L_{1}})}^{2} = (1.002)^{2}$
$R_{2} = 1.004 R_{1}$
$R_{2}$ is increased by $0.4 %$ of $R_{1}$

AMU-2017

Current Electricity

151910 Six wires, each of resistance $r$ , are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

1

r

2

2 r

3

\frac{r}{3}

4

\frac{r}{2}

Explanation:

D
Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit.
$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r}$
$R = r$
It will in parallel with outer resistance
$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$
$R_{eq} = \frac{r}{2}$

WB JEE 2017

Current Electricity

151904 An electric motor operates on a $50 V$ supply and a current of $12 A$ . If the efficiency of the motor is $30 %$ , what is the resistance of the winding of the motor?

1

6 Ω

2

4 Ω

3

2.9 Ω

4

3.1 Ω

Explanation:

C Given,
$V = 50 Volt$
$I = 12 A$
Efficiency $(η) = 30 %$
Power Input $(P_{I}) = 50 \times 12$
$= 600 W$
Power output $(P_{o}) = \frac{30}{100} \times 600 = 180 W$
Net power, $P = (P_{I} - P_{o})$
$= 600 - 180 = 420 W$
We know that,
Power, $P = I^{2} R$
$420 = (12)^{2} \times R$
$R = \frac{420}{144} = 2.9 Ω$

Manipal UGET-2018

Current Electricity

151905 A uniform wire of resistance $9 Ω$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC$ . A cell of emf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1

1 V

2

2 V

3

3 V

4

0.5 V

Explanation:

A
$∵ 3 Ω$ and $3 Ω$ are series

In the circuit,
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = 2 Ω$
$R_{eq} = 2 Ω$
Now,
$V = IR$
$2 = I \times 2$
$I = 1 A$
$∴ i_{1} = 1 \times (\frac{3}{3 + 6})$
$i_{1} = \frac{1}{3} A$
Potential difference between $A$ and $B$ ,
$V_{A} - V_{B} = \frac{1}{3} \times R$
$V_{A} - V_{B} = \frac{1}{3} \times 3$
$V_{A} - V_{B} = 1 V$

Manipal UGET-2018

Current Electricity

151907 A circular coil of single turn has a resistance of $20 Ω$ . Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?

1

5 Ω

2

10 Ω

3

20 Ω

4

40 Ω

Explanation:

A Given, Resistance of circular coil of single turn $= 20 Ω$
If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel.

$∴$ Equivalent Resistance,
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10}$
$∴ R_{eq} = 5 Ω$

NDA (I) 2017

Current Electricity

151908 On stretching a wire its length is increased by $0.2 %$ , its resistance will

1 increase by

0.1 %

2 decrease by

0.1 %

3 increase by

0.2 %

4 increase by

0.4 %

Explanation:

D $L_{2} = L_{1} + 0.2 %$ of $L_{1}$
$L_{2} = 1.002 L_{1}$
$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{V}$
Where $A$ is area, and $V$ is volume
$R \propto L^{2}$
Current Electricity
$\frac{R_{2}}{R_{1}} = {(\frac{L_{2}}{L_{1}})}^{2}$
$\frac{R_{2}}{R_{1}} = {(\frac{1.002 L_{1}}{L_{1}})}^{2} = (1.002)^{2}$
$R_{2} = 1.004 R_{1}$
$R_{2}$ is increased by $0.4 %$ of $R_{1}$

AMU-2017

Current Electricity

151910 Six wires, each of resistance $r$ , are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

1

r

2

2 r

3

\frac{r}{3}

4

\frac{r}{2}

Explanation:

D
Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit.
$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r}$
$R = r$
It will in parallel with outer resistance
$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$
$R_{eq} = \frac{r}{2}$

WB JEE 2017

Current Electricity

151904 An electric motor operates on a $50 V$ supply and a current of $12 A$ . If the efficiency of the motor is $30 %$ , what is the resistance of the winding of the motor?

1

6 Ω

2

4 Ω

3

2.9 Ω

4

3.1 Ω

Explanation:

C Given,
$V = 50 Volt$
$I = 12 A$
Efficiency $(η) = 30 %$
Power Input $(P_{I}) = 50 \times 12$
$= 600 W$
Power output $(P_{o}) = \frac{30}{100} \times 600 = 180 W$
Net power, $P = (P_{I} - P_{o})$
$= 600 - 180 = 420 W$
We know that,
Power, $P = I^{2} R$
$420 = (12)^{2} \times R$
$R = \frac{420}{144} = 2.9 Ω$

Manipal UGET-2018

Current Electricity

151905 A uniform wire of resistance $9 Ω$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC$ . A cell of emf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1

1 V

2

2 V

3

3 V

4

0.5 V

Explanation:

A
$∵ 3 Ω$ and $3 Ω$ are series

In the circuit,
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = 2 Ω$
$R_{eq} = 2 Ω$
Now,
$V = IR$
$2 = I \times 2$
$I = 1 A$
$∴ i_{1} = 1 \times (\frac{3}{3 + 6})$
$i_{1} = \frac{1}{3} A$
Potential difference between $A$ and $B$ ,
$V_{A} - V_{B} = \frac{1}{3} \times R$
$V_{A} - V_{B} = \frac{1}{3} \times 3$
$V_{A} - V_{B} = 1 V$

Manipal UGET-2018

Current Electricity

151907 A circular coil of single turn has a resistance of $20 Ω$ . Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?

1

5 Ω

2

10 Ω

3

20 Ω

4

40 Ω

Explanation:

A Given, Resistance of circular coil of single turn $= 20 Ω$
If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel.

$∴$ Equivalent Resistance,
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10}$
$∴ R_{eq} = 5 Ω$

NDA (I) 2017

Current Electricity

151908 On stretching a wire its length is increased by $0.2 %$ , its resistance will

1 increase by

0.1 %

2 decrease by

0.1 %

3 increase by

0.2 %

4 increase by

0.4 %

Explanation:

D $L_{2} = L_{1} + 0.2 %$ of $L_{1}$
$L_{2} = 1.002 L_{1}$
$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{V}$
Where $A$ is area, and $V$ is volume
$R \propto L^{2}$
Current Electricity
$\frac{R_{2}}{R_{1}} = {(\frac{L_{2}}{L_{1}})}^{2}$
$\frac{R_{2}}{R_{1}} = {(\frac{1.002 L_{1}}{L_{1}})}^{2} = (1.002)^{2}$
$R_{2} = 1.004 R_{1}$
$R_{2}$ is increased by $0.4 %$ of $R_{1}$

AMU-2017

Current Electricity

151910 Six wires, each of resistance $r$ , are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

1

r

2

2 r

3

\frac{r}{3}

4

\frac{r}{2}

Explanation:

D
Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit.
$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r}$
$R = r$
It will in parallel with outer resistance
$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$
$R_{eq} = \frac{r}{2}$

WB JEE 2017

Current Electricity

151904 An electric motor operates on a $50 V$ supply and a current of $12 A$ . If the efficiency of the motor is $30 %$ , what is the resistance of the winding of the motor?

1

6 Ω

2

4 Ω

3

2.9 Ω

4

3.1 Ω

Explanation:

C Given,
$V = 50 Volt$
$I = 12 A$
Efficiency $(η) = 30 %$
Power Input $(P_{I}) = 50 \times 12$
$= 600 W$
Power output $(P_{o}) = \frac{30}{100} \times 600 = 180 W$
Net power, $P = (P_{I} - P_{o})$
$= 600 - 180 = 420 W$
We know that,
Power, $P = I^{2} R$
$420 = (12)^{2} \times R$
$R = \frac{420}{144} = 2.9 Ω$

Manipal UGET-2018

Current Electricity

151905 A uniform wire of resistance $9 Ω$ is cut into 3 equal parts. They are connected in the form of equilateral triangle $ABC$ . A cell of emf $2 V$ and negligible internal resistance is connected across $B$ and $C$ . Potential difference across $A B$ is

1

1 V

2

2 V

3

3 V

4

0.5 V

Explanation:

A
$∵ 3 Ω$ and $3 Ω$ are series

In the circuit,
$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = 2 Ω$
$R_{eq} = 2 Ω$
Now,
$V = IR$
$2 = I \times 2$
$I = 1 A$
$∴ i_{1} = 1 \times (\frac{3}{3 + 6})$
$i_{1} = \frac{1}{3} A$
Potential difference between $A$ and $B$ ,
$V_{A} - V_{B} = \frac{1}{3} \times R$
$V_{A} - V_{B} = \frac{1}{3} \times 3$
$V_{A} - V_{B} = 1 V$

Manipal UGET-2018

Current Electricity

151907 A circular coil of single turn has a resistance of $20 Ω$ . Which one of the following is the correct value for the resistance between the ends of any diameter of the coil?

1

5 Ω

2

10 Ω

3

20 Ω

4

40 Ω

Explanation:

A Given, Resistance of circular coil of single turn $= 20 Ω$
If we measure the resistance across the diameter, it will be like two semicircular loops connected in parallel.

$∴$ Equivalent Resistance,
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10}$
$∴ R_{eq} = 5 Ω$

NDA (I) 2017

Current Electricity

151908 On stretching a wire its length is increased by $0.2 %$ , its resistance will

1 increase by

0.1 %

2 decrease by

0.1 %

3 increase by

0.2 %

4 increase by

0.4 %

Explanation:

D $L_{2} = L_{1} + 0.2 %$ of $L_{1}$
$L_{2} = 1.002 L_{1}$
$R = ρ \frac{L}{A} = ρ \frac{L^{2}}{V}$
Where $A$ is area, and $V$ is volume
$R \propto L^{2}$
Current Electricity
$\frac{R_{2}}{R_{1}} = {(\frac{L_{2}}{L_{1}})}^{2}$
$\frac{R_{2}}{R_{1}} = {(\frac{1.002 L_{1}}{L_{1}})}^{2} = (1.002)^{2}$
$R_{2} = 1.004 R_{1}$
$R_{2}$ is increased by $0.4 %$ of $R_{1}$

AMU-2017

Current Electricity

151910 Six wires, each of resistance $r$ , are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

1

r

2

2 r

3

\frac{r}{3}

4

\frac{r}{2}

Explanation:

D
Six wires each of resistance $r$ from a tetrahedron unit. The equivalent circuit of this tetrahedron. Hence it is circuit of Wheatstone the equivalent resistance of upper circuit.
$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} = \frac{2}{2 r}$
$R = r$
It will in parallel with outer resistance
$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$
$R_{eq} = \frac{r}{2}$

WB JEE 2017