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PHXII06:ELECTROMAGNETIC INDUCTION

358362 Two inductors $L_{1}$ and $L_{2}$ are connected in parallel and a time varying current flows as shown. The ratio $\frac{i_{1}}{i_{2}}$ is equal to
supporting img

1

L_{2} / L_{1}

2

L_{1} / L_{2}

3

\frac{L_{2}^{2}}{{(L_{1} + L_{2})}^{2}}

4

\frac{L_{1}^{2}}{{(L_{1} + L_{2})}^{2}}

Explanation:

The inductors are in parallel. Therefore, potential difference across them is same.
$L_{1} (\frac{d i_{1}}{d t}) = L_{2} (\frac{d i_{2}}{d t})$
or $L_{1} (d i_{1}) = L_{2} (d i_{2})$
Integrating, we get
$L_{1} i_{1} = L_{2} i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

PHXII06:ELECTROMAGNETIC INDUCTION

358363 The equivalent inductance of two inductances is 2.4 henry when connected in parallel and 10 henry when connected in series. The difference between the two inductances is

1

2 h e n r y

2

5 h e n r y

3

3 h e n r y

4

4 h e n r y

Explanation:

$L_{S} = L_{1} + L_{2} = 10 H (1)$
$L_{P} = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2.4 H (2)$
On solving (1) and (2) $L_{1} L_{2} = 24 (3)$
Also ${(L_{1} - L_{2})}^{2} = {(L_{1} + L_{2})}^{2} - 4 L_{1} L_{2}$
$\Rightarrow {(L_{1} - L_{2})}^{2} = (10)^{2} - 4 \times 24 = 4 \Rightarrow L_{1} - L_{2} = 2 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358364 Two identical induction coils each of inductance $L$ joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other. What is the net inductance?

1

L^{2}

2

2 L

3

L / 2

4 zero

Explanation:

When the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is $180^{\circ}$ out of phase of the emf produced in second coil.
$\begin{aligned} L_{net} & = L_{1} + L_{2} \\ = - \frac{ϕ}{i} + \frac{ϕ}{i} = 0 \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358365 Pure inductance of $3.0 H$ is connected as shown below. The equivalent inductance of the circuit is
supporting img

1

9 H

2

3 H

3

2 H

4

1 H

Explanation:

The inductances are in parallel
$\Rightarrow L_{e q} = \frac{L}{3} = \frac{3}{3} = 1 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358362 Two inductors $L_{1}$ and $L_{2}$ are connected in parallel and a time varying current flows as shown. The ratio $\frac{i_{1}}{i_{2}}$ is equal to
supporting img

1

L_{2} / L_{1}

2

L_{1} / L_{2}

3

\frac{L_{2}^{2}}{{(L_{1} + L_{2})}^{2}}

4

\frac{L_{1}^{2}}{{(L_{1} + L_{2})}^{2}}

Explanation:

The inductors are in parallel. Therefore, potential difference across them is same.
$L_{1} (\frac{d i_{1}}{d t}) = L_{2} (\frac{d i_{2}}{d t})$
or $L_{1} (d i_{1}) = L_{2} (d i_{2})$
Integrating, we get
$L_{1} i_{1} = L_{2} i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

PHXII06:ELECTROMAGNETIC INDUCTION

358363 The equivalent inductance of two inductances is 2.4 henry when connected in parallel and 10 henry when connected in series. The difference between the two inductances is

1

2 h e n r y

2

5 h e n r y

3

3 h e n r y

4

4 h e n r y

Explanation:

$L_{S} = L_{1} + L_{2} = 10 H (1)$
$L_{P} = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2.4 H (2)$
On solving (1) and (2) $L_{1} L_{2} = 24 (3)$
Also ${(L_{1} - L_{2})}^{2} = {(L_{1} + L_{2})}^{2} - 4 L_{1} L_{2}$
$\Rightarrow {(L_{1} - L_{2})}^{2} = (10)^{2} - 4 \times 24 = 4 \Rightarrow L_{1} - L_{2} = 2 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358364 Two identical induction coils each of inductance $L$ joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other. What is the net inductance?

1

L^{2}

2

2 L

3

L / 2

4 zero

Explanation:

When the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is $180^{\circ}$ out of phase of the emf produced in second coil.
$\begin{aligned} L_{net} & = L_{1} + L_{2} \\ = - \frac{ϕ}{i} + \frac{ϕ}{i} = 0 \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358365 Pure inductance of $3.0 H$ is connected as shown below. The equivalent inductance of the circuit is
supporting img

1

9 H

2

3 H

3

2 H

4

1 H

Explanation:

The inductances are in parallel
$\Rightarrow L_{e q} = \frac{L}{3} = \frac{3}{3} = 1 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358362 Two inductors $L_{1}$ and $L_{2}$ are connected in parallel and a time varying current flows as shown. The ratio $\frac{i_{1}}{i_{2}}$ is equal to
supporting img

1

L_{2} / L_{1}

2

L_{1} / L_{2}

3

\frac{L_{2}^{2}}{{(L_{1} + L_{2})}^{2}}

4

\frac{L_{1}^{2}}{{(L_{1} + L_{2})}^{2}}

Explanation:

The inductors are in parallel. Therefore, potential difference across them is same.
$L_{1} (\frac{d i_{1}}{d t}) = L_{2} (\frac{d i_{2}}{d t})$
or $L_{1} (d i_{1}) = L_{2} (d i_{2})$
Integrating, we get
$L_{1} i_{1} = L_{2} i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

PHXII06:ELECTROMAGNETIC INDUCTION

358363 The equivalent inductance of two inductances is 2.4 henry when connected in parallel and 10 henry when connected in series. The difference between the two inductances is

1

2 h e n r y

2

5 h e n r y

3

3 h e n r y

4

4 h e n r y

Explanation:

$L_{S} = L_{1} + L_{2} = 10 H (1)$
$L_{P} = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2.4 H (2)$
On solving (1) and (2) $L_{1} L_{2} = 24 (3)$
Also ${(L_{1} - L_{2})}^{2} = {(L_{1} + L_{2})}^{2} - 4 L_{1} L_{2}$
$\Rightarrow {(L_{1} - L_{2})}^{2} = (10)^{2} - 4 \times 24 = 4 \Rightarrow L_{1} - L_{2} = 2 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358364 Two identical induction coils each of inductance $L$ joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other. What is the net inductance?

1

L^{2}

2

2 L

3

L / 2

4 zero

Explanation:

When the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is $180^{\circ}$ out of phase of the emf produced in second coil.
$\begin{aligned} L_{net} & = L_{1} + L_{2} \\ = - \frac{ϕ}{i} + \frac{ϕ}{i} = 0 \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358365 Pure inductance of $3.0 H$ is connected as shown below. The equivalent inductance of the circuit is
supporting img

1

9 H

2

3 H

3

2 H

4

1 H

Explanation:

The inductances are in parallel
$\Rightarrow L_{e q} = \frac{L}{3} = \frac{3}{3} = 1 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358362 Two inductors $L_{1}$ and $L_{2}$ are connected in parallel and a time varying current flows as shown. The ratio $\frac{i_{1}}{i_{2}}$ is equal to
supporting img

1

L_{2} / L_{1}

2

L_{1} / L_{2}

3

\frac{L_{2}^{2}}{{(L_{1} + L_{2})}^{2}}

4

\frac{L_{1}^{2}}{{(L_{1} + L_{2})}^{2}}

Explanation:

The inductors are in parallel. Therefore, potential difference across them is same.
$L_{1} (\frac{d i_{1}}{d t}) = L_{2} (\frac{d i_{2}}{d t})$
or $L_{1} (d i_{1}) = L_{2} (d i_{2})$
Integrating, we get
$L_{1} i_{1} = L_{2} i_{2} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

PHXII06:ELECTROMAGNETIC INDUCTION

358363 The equivalent inductance of two inductances is 2.4 henry when connected in parallel and 10 henry when connected in series. The difference between the two inductances is

1

2 h e n r y

2

5 h e n r y

3

3 h e n r y

4

4 h e n r y

Explanation:

$L_{S} = L_{1} + L_{2} = 10 H (1)$
$L_{P} = \frac{L_{1} L_{2}}{L_{1} + L_{2}} = 2.4 H (2)$
On solving (1) and (2) $L_{1} L_{2} = 24 (3)$
Also ${(L_{1} - L_{2})}^{2} = {(L_{1} + L_{2})}^{2} - 4 L_{1} L_{2}$
$\Rightarrow {(L_{1} - L_{2})}^{2} = (10)^{2} - 4 \times 24 = 4 \Rightarrow L_{1} - L_{2} = 2 H$

PHXII06:ELECTROMAGNETIC INDUCTION

358364 Two identical induction coils each of inductance $L$ joined in series are placed very close to each other such that the winding direction of one is exactly opposite to that of the other. What is the net inductance?

1

L^{2}

2

2 L

3

L / 2

4 zero

Explanation:

When the two coils are joined in series such that the winding of one is opposite to the other, then the emf produced in first coil is $180^{\circ}$ out of phase of the emf produced in second coil.
$\begin{aligned} L_{net} & = L_{1} + L_{2} \\ = - \frac{ϕ}{i} + \frac{ϕ}{i} = 0 \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358365 Pure inductance of $3.0 H$ is connected as shown below. The equivalent inductance of the circuit is
supporting img

1

9 H

2

3 H

3

2 H

4

1 H

Explanation:

The inductances are in parallel
$\Rightarrow L_{e q} = \frac{L}{3} = \frac{3}{3} = 1 H$