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Electro Magnetic Induction

154739 The currents in two coils of self- inductance 5 $mH$ and $1 mH$ are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following

1 The ratio of induced emf 's is

5 : 1

2 The ratio of current in the coils is

1 : 5

3 The ratio of energy stored in the two coils at that instant is

1 : 5

4 The ratio of energy stored in the two coils at the instant is

9 : 25

Explanation:

D Given that,
$L_{1} = 5 mH, L_{2} = 1 mH$
$\frac{dI}{dt}$ is constant in both coils
Power supplied ' $P$ ' is also same.
Ratio of induced emf -
$\frac{e_{1}}{e_{2}} = \frac{L_{1} dI / dt}{L_{2} dI / dt} = \frac{5}{1}$
Ratio of current -
$∵$ Power is same - i.e. is constant
$P_{1} = P_{2}$
$e_{1} i_{1} = e_{2} i_{2}$
$\frac{i_{1}}{i_{2}} = \frac{e_{2}}{e_{1}}$
Form equation (i), we get-
$\frac{i_{1}}{i_{2}} = \frac{1}{5}$
Ratio of energy stored -
$\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} L_{1} {(i_{1})}^{2}}{\frac{1}{2} L_{2} {(i_{2})}^{2}} = \frac{5}{1} \times {(\frac{1}{5})}^{2}$
$\frac{E_{1}}{E_{2}} = \frac{1}{5}$
Hence, option (d) is false statement.

AP EAMCET (Medical)-05.10.2021

Electro Magnetic Induction

154741 A magnetic field given by $B (t) = (0.2 t - 0.05$ $t^{2}) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 $cm$ and whose total resistance is $5 Ω$ . The power dissipation at 3 seconds is nearly

1

4 μ W

2

7 μ W

3

2.3 μ W

4

1.25 μ W

5

1.29 μ W

Explanation:

E Given that,
$B (t) = (0.2 t - 0.05 t^{2})$ tesla
$N = 25$ turns, $r = 1.8 cm, R = 5 Ω$
$t = 3 \sec$
Induced emf $(ε) = - N \frac{d ϕ}{dt}$
$ε = - 25 \frac{d}{dt} (B.A)$
$ε = - 25 \times π r^{2} \times \frac{d}{dt} (0.2 t - 0.05 t^{2})$
$ε = - 25 \times π \times (1.8)^{2} \times 10^{- 4} \times [0.2 - 0.1 t]$
At, $t = 3 \sec$
$ε = - 254.5 \times 10^{- 4} [0.2 - 0.1 \times 3]$
$ε = 0.254 \times 10^{- 2}$
We know that,
Power dissipated is given by -
$P = \frac{ε^{2}}{R} = \frac{{(0.254 \times 10^{- 2})}^{2}}{5}$
$P = \frac{0.064516}{5} \times 10^{- 4} = 0.0129 \times 10^{- 4}$
$P = 1.29 μ W$

AP EAMCET-24.08.2021

Electro Magnetic Induction

154742 Equivalent inductance of two inductors is $2.4 H$ when connected in parallel and $10 H$ when connected in series. What is the value of inductance of the individual inductors?

1

8 H & 2 H

2

6 H & 4 H

3

5 H & 5 H

4

7 H & 3 H

Explanation:

B Given,
$L_{1} + L_{2} = 10$
(connected in series)
And
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{2.4}$
(connected in parallel)
Also
$\frac{L_{1} \cdot L_{2}}{L_{1} + L_{2}} = 2.4$
$\frac{L_{1} \cdot (10 - L_{1})}{L_{1} + (10 - L_{1})} = 2.4$
$10 L_{1} - L_{1}^{2} = 24$
$L_{1}^{2} - 10 L_{1} + 24 = 0$
$L_{1}^{2} - 6 L_{1} - 4 L_{1} + 24 = 0$
$L_{1} (L_{1} - 6) - 4 (L_{1} - 6) = 0$
$L_{1} = 6, 4$
The two inductors are $6 H$ and $4 H$ .

AP EAMCET-03.09.2021

Electro Magnetic Induction

154743 Find the equivalent inductance between $A$ and $B$ in the given circuit.

1

1 H

2

4 H

3

0.8 H

4

16 H

Explanation:

A The diagram can be resolved as -

The combination is parallel -
Hence, $\frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} + \frac{1}{L_{4}}$
$\frac{1}{L_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4}$
$L_{eq} = 1 H$

AP EAMCET-03.09.2021

Electro Magnetic Induction

154744 Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 H$ , and when connected in series are equivalent to a single inductor of inductance $8 H$ . Find the difference in the inductances of $A$ and $B$ .

1

3 H

2

7.5 H

3

2 H

4

4 H

Explanation:

D Given,
$L_{1} + L_{2} = 8 H \dots . (i) (connected in series)$
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{1.5} \dots . (ii) (connected in parallel)$
Also, $\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5$
$\frac{L_{1} (8 - L_{1})}{L_{1} + (8 - L_{1})} = 1.5$
$8 L_{1} - L_{1}^{2} = 12$
$L_{1}^{2} - 8 L_{1} + 12 = 0$
$L_{1}^{2} - 6 L_{1} - 2 L_{1} + 12 = 0$
$L_{1} (L_{1} - 6) - 2 (L_{1} - 6) = 0$
$L = 2, 6$
The difference between inductances $A$ and $B$ is $= 6 - 2 = 4 H$

AP EAMCET-19.08.2021

Electro Magnetic Induction

154739 The currents in two coils of self- inductance 5 $mH$ and $1 mH$ are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following

1 The ratio of induced emf 's is

5 : 1

2 The ratio of current in the coils is

1 : 5

3 The ratio of energy stored in the two coils at that instant is

1 : 5

4 The ratio of energy stored in the two coils at the instant is

9 : 25

Explanation:

D Given that,
$L_{1} = 5 mH, L_{2} = 1 mH$
$\frac{dI}{dt}$ is constant in both coils
Power supplied ' $P$ ' is also same.
Ratio of induced emf -
$\frac{e_{1}}{e_{2}} = \frac{L_{1} dI / dt}{L_{2} dI / dt} = \frac{5}{1}$
Ratio of current -
$∵$ Power is same - i.e. is constant
$P_{1} = P_{2}$
$e_{1} i_{1} = e_{2} i_{2}$
$\frac{i_{1}}{i_{2}} = \frac{e_{2}}{e_{1}}$
Form equation (i), we get-
$\frac{i_{1}}{i_{2}} = \frac{1}{5}$
Ratio of energy stored -
$\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} L_{1} {(i_{1})}^{2}}{\frac{1}{2} L_{2} {(i_{2})}^{2}} = \frac{5}{1} \times {(\frac{1}{5})}^{2}$
$\frac{E_{1}}{E_{2}} = \frac{1}{5}$
Hence, option (d) is false statement.

AP EAMCET (Medical)-05.10.2021

Electro Magnetic Induction

154741 A magnetic field given by $B (t) = (0.2 t - 0.05$ $t^{2}) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 $cm$ and whose total resistance is $5 Ω$ . The power dissipation at 3 seconds is nearly

1

4 μ W

2

7 μ W

3

2.3 μ W

4

1.25 μ W

5

1.29 μ W

Explanation:

E Given that,
$B (t) = (0.2 t - 0.05 t^{2})$ tesla
$N = 25$ turns, $r = 1.8 cm, R = 5 Ω$
$t = 3 \sec$
Induced emf $(ε) = - N \frac{d ϕ}{dt}$
$ε = - 25 \frac{d}{dt} (B.A)$
$ε = - 25 \times π r^{2} \times \frac{d}{dt} (0.2 t - 0.05 t^{2})$
$ε = - 25 \times π \times (1.8)^{2} \times 10^{- 4} \times [0.2 - 0.1 t]$
At, $t = 3 \sec$
$ε = - 254.5 \times 10^{- 4} [0.2 - 0.1 \times 3]$
$ε = 0.254 \times 10^{- 2}$
We know that,
Power dissipated is given by -
$P = \frac{ε^{2}}{R} = \frac{{(0.254 \times 10^{- 2})}^{2}}{5}$
$P = \frac{0.064516}{5} \times 10^{- 4} = 0.0129 \times 10^{- 4}$
$P = 1.29 μ W$

AP EAMCET-24.08.2021

Electro Magnetic Induction

154742 Equivalent inductance of two inductors is $2.4 H$ when connected in parallel and $10 H$ when connected in series. What is the value of inductance of the individual inductors?

1

8 H & 2 H

2

6 H & 4 H

3

5 H & 5 H

4

7 H & 3 H

Explanation:

B Given,
$L_{1} + L_{2} = 10$
(connected in series)
And
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{2.4}$
(connected in parallel)
Also
$\frac{L_{1} \cdot L_{2}}{L_{1} + L_{2}} = 2.4$
$\frac{L_{1} \cdot (10 - L_{1})}{L_{1} + (10 - L_{1})} = 2.4$
$10 L_{1} - L_{1}^{2} = 24$
$L_{1}^{2} - 10 L_{1} + 24 = 0$
$L_{1}^{2} - 6 L_{1} - 4 L_{1} + 24 = 0$
$L_{1} (L_{1} - 6) - 4 (L_{1} - 6) = 0$
$L_{1} = 6, 4$
The two inductors are $6 H$ and $4 H$ .

AP EAMCET-03.09.2021

Electro Magnetic Induction

154743 Find the equivalent inductance between $A$ and $B$ in the given circuit.

1

1 H

2

4 H

3

0.8 H

4

16 H

Explanation:

A The diagram can be resolved as -

The combination is parallel -
Hence, $\frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} + \frac{1}{L_{4}}$
$\frac{1}{L_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4}$
$L_{eq} = 1 H$

AP EAMCET-03.09.2021

Electro Magnetic Induction

154744 Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 H$ , and when connected in series are equivalent to a single inductor of inductance $8 H$ . Find the difference in the inductances of $A$ and $B$ .

1

3 H

2

7.5 H

3

2 H

4

4 H

Explanation:

D Given,
$L_{1} + L_{2} = 8 H \dots . (i) (connected in series)$
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{1.5} \dots . (ii) (connected in parallel)$
Also, $\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5$
$\frac{L_{1} (8 - L_{1})}{L_{1} + (8 - L_{1})} = 1.5$
$8 L_{1} - L_{1}^{2} = 12$
$L_{1}^{2} - 8 L_{1} + 12 = 0$
$L_{1}^{2} - 6 L_{1} - 2 L_{1} + 12 = 0$
$L_{1} (L_{1} - 6) - 2 (L_{1} - 6) = 0$
$L = 2, 6$
The difference between inductances $A$ and $B$ is $= 6 - 2 = 4 H$

AP EAMCET-19.08.2021

Electro Magnetic Induction

154739 The currents in two coils of self- inductance 5 $mH$ and $1 mH$ are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following

1 The ratio of induced emf 's is

5 : 1

2 The ratio of current in the coils is

1 : 5

3 The ratio of energy stored in the two coils at that instant is

1 : 5

4 The ratio of energy stored in the two coils at the instant is

9 : 25

Explanation:

D Given that,
$L_{1} = 5 mH, L_{2} = 1 mH$
$\frac{dI}{dt}$ is constant in both coils
Power supplied ' $P$ ' is also same.
Ratio of induced emf -
$\frac{e_{1}}{e_{2}} = \frac{L_{1} dI / dt}{L_{2} dI / dt} = \frac{5}{1}$
Ratio of current -
$∵$ Power is same - i.e. is constant
$P_{1} = P_{2}$
$e_{1} i_{1} = e_{2} i_{2}$
$\frac{i_{1}}{i_{2}} = \frac{e_{2}}{e_{1}}$
Form equation (i), we get-
$\frac{i_{1}}{i_{2}} = \frac{1}{5}$
Ratio of energy stored -
$\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} L_{1} {(i_{1})}^{2}}{\frac{1}{2} L_{2} {(i_{2})}^{2}} = \frac{5}{1} \times {(\frac{1}{5})}^{2}$
$\frac{E_{1}}{E_{2}} = \frac{1}{5}$
Hence, option (d) is false statement.

AP EAMCET (Medical)-05.10.2021

Electro Magnetic Induction

154741 A magnetic field given by $B (t) = (0.2 t - 0.05$ $t^{2}) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 $cm$ and whose total resistance is $5 Ω$ . The power dissipation at 3 seconds is nearly

1

4 μ W

2

7 μ W

3

2.3 μ W

4

1.25 μ W

5

1.29 μ W

Explanation:

E Given that,
$B (t) = (0.2 t - 0.05 t^{2})$ tesla
$N = 25$ turns, $r = 1.8 cm, R = 5 Ω$
$t = 3 \sec$
Induced emf $(ε) = - N \frac{d ϕ}{dt}$
$ε = - 25 \frac{d}{dt} (B.A)$
$ε = - 25 \times π r^{2} \times \frac{d}{dt} (0.2 t - 0.05 t^{2})$
$ε = - 25 \times π \times (1.8)^{2} \times 10^{- 4} \times [0.2 - 0.1 t]$
At, $t = 3 \sec$
$ε = - 254.5 \times 10^{- 4} [0.2 - 0.1 \times 3]$
$ε = 0.254 \times 10^{- 2}$
We know that,
Power dissipated is given by -
$P = \frac{ε^{2}}{R} = \frac{{(0.254 \times 10^{- 2})}^{2}}{5}$
$P = \frac{0.064516}{5} \times 10^{- 4} = 0.0129 \times 10^{- 4}$
$P = 1.29 μ W$

AP EAMCET-24.08.2021

Electro Magnetic Induction

154742 Equivalent inductance of two inductors is $2.4 H$ when connected in parallel and $10 H$ when connected in series. What is the value of inductance of the individual inductors?

1

8 H & 2 H

2

6 H & 4 H

3

5 H & 5 H

4

7 H & 3 H

Explanation:

B Given,
$L_{1} + L_{2} = 10$
(connected in series)
And
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{2.4}$
(connected in parallel)
Also
$\frac{L_{1} \cdot L_{2}}{L_{1} + L_{2}} = 2.4$
$\frac{L_{1} \cdot (10 - L_{1})}{L_{1} + (10 - L_{1})} = 2.4$
$10 L_{1} - L_{1}^{2} = 24$
$L_{1}^{2} - 10 L_{1} + 24 = 0$
$L_{1}^{2} - 6 L_{1} - 4 L_{1} + 24 = 0$
$L_{1} (L_{1} - 6) - 4 (L_{1} - 6) = 0$
$L_{1} = 6, 4$
The two inductors are $6 H$ and $4 H$ .

AP EAMCET-03.09.2021

Electro Magnetic Induction

154743 Find the equivalent inductance between $A$ and $B$ in the given circuit.

1

1 H

2

4 H

3

0.8 H

4

16 H

Explanation:

A The diagram can be resolved as -

The combination is parallel -
Hence, $\frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} + \frac{1}{L_{4}}$
$\frac{1}{L_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4}$
$L_{eq} = 1 H$

AP EAMCET-03.09.2021

Electro Magnetic Induction

154744 Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 H$ , and when connected in series are equivalent to a single inductor of inductance $8 H$ . Find the difference in the inductances of $A$ and $B$ .

1

3 H

2

7.5 H

3

2 H

4

4 H

Explanation:

D Given,
$L_{1} + L_{2} = 8 H \dots . (i) (connected in series)$
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{1.5} \dots . (ii) (connected in parallel)$
Also, $\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5$
$\frac{L_{1} (8 - L_{1})}{L_{1} + (8 - L_{1})} = 1.5$
$8 L_{1} - L_{1}^{2} = 12$
$L_{1}^{2} - 8 L_{1} + 12 = 0$
$L_{1}^{2} - 6 L_{1} - 2 L_{1} + 12 = 0$
$L_{1} (L_{1} - 6) - 2 (L_{1} - 6) = 0$
$L = 2, 6$
The difference between inductances $A$ and $B$ is $= 6 - 2 = 4 H$

AP EAMCET-19.08.2021

Electro Magnetic Induction

154739 The currents in two coils of self- inductance 5 $mH$ and $1 mH$ are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following

1 The ratio of induced emf 's is

5 : 1

2 The ratio of current in the coils is

1 : 5

3 The ratio of energy stored in the two coils at that instant is

1 : 5

4 The ratio of energy stored in the two coils at the instant is

9 : 25

Explanation:

D Given that,
$L_{1} = 5 mH, L_{2} = 1 mH$
$\frac{dI}{dt}$ is constant in both coils
Power supplied ' $P$ ' is also same.
Ratio of induced emf -
$\frac{e_{1}}{e_{2}} = \frac{L_{1} dI / dt}{L_{2} dI / dt} = \frac{5}{1}$
Ratio of current -
$∵$ Power is same - i.e. is constant
$P_{1} = P_{2}$
$e_{1} i_{1} = e_{2} i_{2}$
$\frac{i_{1}}{i_{2}} = \frac{e_{2}}{e_{1}}$
Form equation (i), we get-
$\frac{i_{1}}{i_{2}} = \frac{1}{5}$
Ratio of energy stored -
$\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} L_{1} {(i_{1})}^{2}}{\frac{1}{2} L_{2} {(i_{2})}^{2}} = \frac{5}{1} \times {(\frac{1}{5})}^{2}$
$\frac{E_{1}}{E_{2}} = \frac{1}{5}$
Hence, option (d) is false statement.

AP EAMCET (Medical)-05.10.2021

Electro Magnetic Induction

154741 A magnetic field given by $B (t) = (0.2 t - 0.05$ $t^{2}) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 $cm$ and whose total resistance is $5 Ω$ . The power dissipation at 3 seconds is nearly

1

4 μ W

2

7 μ W

3

2.3 μ W

4

1.25 μ W

5

1.29 μ W

Explanation:

E Given that,
$B (t) = (0.2 t - 0.05 t^{2})$ tesla
$N = 25$ turns, $r = 1.8 cm, R = 5 Ω$
$t = 3 \sec$
Induced emf $(ε) = - N \frac{d ϕ}{dt}$
$ε = - 25 \frac{d}{dt} (B.A)$
$ε = - 25 \times π r^{2} \times \frac{d}{dt} (0.2 t - 0.05 t^{2})$
$ε = - 25 \times π \times (1.8)^{2} \times 10^{- 4} \times [0.2 - 0.1 t]$
At, $t = 3 \sec$
$ε = - 254.5 \times 10^{- 4} [0.2 - 0.1 \times 3]$
$ε = 0.254 \times 10^{- 2}$
We know that,
Power dissipated is given by -
$P = \frac{ε^{2}}{R} = \frac{{(0.254 \times 10^{- 2})}^{2}}{5}$
$P = \frac{0.064516}{5} \times 10^{- 4} = 0.0129 \times 10^{- 4}$
$P = 1.29 μ W$

AP EAMCET-24.08.2021

Electro Magnetic Induction

154742 Equivalent inductance of two inductors is $2.4 H$ when connected in parallel and $10 H$ when connected in series. What is the value of inductance of the individual inductors?

1

8 H & 2 H

2

6 H & 4 H

3

5 H & 5 H

4

7 H & 3 H

Explanation:

B Given,
$L_{1} + L_{2} = 10$
(connected in series)
And
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{2.4}$
(connected in parallel)
Also
$\frac{L_{1} \cdot L_{2}}{L_{1} + L_{2}} = 2.4$
$\frac{L_{1} \cdot (10 - L_{1})}{L_{1} + (10 - L_{1})} = 2.4$
$10 L_{1} - L_{1}^{2} = 24$
$L_{1}^{2} - 10 L_{1} + 24 = 0$
$L_{1}^{2} - 6 L_{1} - 4 L_{1} + 24 = 0$
$L_{1} (L_{1} - 6) - 4 (L_{1} - 6) = 0$
$L_{1} = 6, 4$
The two inductors are $6 H$ and $4 H$ .

AP EAMCET-03.09.2021

Electro Magnetic Induction

154743 Find the equivalent inductance between $A$ and $B$ in the given circuit.

1

1 H

2

4 H

3

0.8 H

4

16 H

Explanation:

A The diagram can be resolved as -

The combination is parallel -
Hence, $\frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} + \frac{1}{L_{4}}$
$\frac{1}{L_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4}$
$L_{eq} = 1 H$

AP EAMCET-03.09.2021

Electro Magnetic Induction

154744 Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 H$ , and when connected in series are equivalent to a single inductor of inductance $8 H$ . Find the difference in the inductances of $A$ and $B$ .

1

3 H

2

7.5 H

3

2 H

4

4 H

Explanation:

D Given,
$L_{1} + L_{2} = 8 H \dots . (i) (connected in series)$
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{1.5} \dots . (ii) (connected in parallel)$
Also, $\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5$
$\frac{L_{1} (8 - L_{1})}{L_{1} + (8 - L_{1})} = 1.5$
$8 L_{1} - L_{1}^{2} = 12$
$L_{1}^{2} - 8 L_{1} + 12 = 0$
$L_{1}^{2} - 6 L_{1} - 2 L_{1} + 12 = 0$
$L_{1} (L_{1} - 6) - 2 (L_{1} - 6) = 0$
$L = 2, 6$
The difference between inductances $A$ and $B$ is $= 6 - 2 = 4 H$

AP EAMCET-19.08.2021

Electro Magnetic Induction

154739 The currents in two coils of self- inductance 5 $mH$ and $1 mH$ are increasing at the same rate at a certain instant. The power supplied to the coils is also same. Then identify the false statement among the following

1 The ratio of induced emf 's is

5 : 1

2 The ratio of current in the coils is

1 : 5

3 The ratio of energy stored in the two coils at that instant is

1 : 5

4 The ratio of energy stored in the two coils at the instant is

9 : 25

Explanation:

D Given that,
$L_{1} = 5 mH, L_{2} = 1 mH$
$\frac{dI}{dt}$ is constant in both coils
Power supplied ' $P$ ' is also same.
Ratio of induced emf -
$\frac{e_{1}}{e_{2}} = \frac{L_{1} dI / dt}{L_{2} dI / dt} = \frac{5}{1}$
Ratio of current -
$∵$ Power is same - i.e. is constant
$P_{1} = P_{2}$
$e_{1} i_{1} = e_{2} i_{2}$
$\frac{i_{1}}{i_{2}} = \frac{e_{2}}{e_{1}}$
Form equation (i), we get-
$\frac{i_{1}}{i_{2}} = \frac{1}{5}$
Ratio of energy stored -
$\frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} L_{1} {(i_{1})}^{2}}{\frac{1}{2} L_{2} {(i_{2})}^{2}} = \frac{5}{1} \times {(\frac{1}{5})}^{2}$
$\frac{E_{1}}{E_{2}} = \frac{1}{5}$
Hence, option (d) is false statement.

AP EAMCET (Medical)-05.10.2021

Electro Magnetic Induction

154741 A magnetic field given by $B (t) = (0.2 t - 0.05$ $t^{2}) T$ is directed perpendicular to the plane of a circular coil containing 25 turns of radius 1.8 $cm$ and whose total resistance is $5 Ω$ . The power dissipation at 3 seconds is nearly

1

4 μ W

2

7 μ W

3

2.3 μ W

4

1.25 μ W

5

1.29 μ W

Explanation:

E Given that,
$B (t) = (0.2 t - 0.05 t^{2})$ tesla
$N = 25$ turns, $r = 1.8 cm, R = 5 Ω$
$t = 3 \sec$
Induced emf $(ε) = - N \frac{d ϕ}{dt}$
$ε = - 25 \frac{d}{dt} (B.A)$
$ε = - 25 \times π r^{2} \times \frac{d}{dt} (0.2 t - 0.05 t^{2})$
$ε = - 25 \times π \times (1.8)^{2} \times 10^{- 4} \times [0.2 - 0.1 t]$
At, $t = 3 \sec$
$ε = - 254.5 \times 10^{- 4} [0.2 - 0.1 \times 3]$
$ε = 0.254 \times 10^{- 2}$
We know that,
Power dissipated is given by -
$P = \frac{ε^{2}}{R} = \frac{{(0.254 \times 10^{- 2})}^{2}}{5}$
$P = \frac{0.064516}{5} \times 10^{- 4} = 0.0129 \times 10^{- 4}$
$P = 1.29 μ W$

AP EAMCET-24.08.2021

Electro Magnetic Induction

154742 Equivalent inductance of two inductors is $2.4 H$ when connected in parallel and $10 H$ when connected in series. What is the value of inductance of the individual inductors?

1

8 H & 2 H

2

6 H & 4 H

3

5 H & 5 H

4

7 H & 3 H

Explanation:

B Given,
$L_{1} + L_{2} = 10$
(connected in series)
And
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{2.4}$
(connected in parallel)
Also
$\frac{L_{1} \cdot L_{2}}{L_{1} + L_{2}} = 2.4$
$\frac{L_{1} \cdot (10 - L_{1})}{L_{1} + (10 - L_{1})} = 2.4$
$10 L_{1} - L_{1}^{2} = 24$
$L_{1}^{2} - 10 L_{1} + 24 = 0$
$L_{1}^{2} - 6 L_{1} - 4 L_{1} + 24 = 0$
$L_{1} (L_{1} - 6) - 4 (L_{1} - 6) = 0$
$L_{1} = 6, 4$
The two inductors are $6 H$ and $4 H$ .

AP EAMCET-03.09.2021

Electro Magnetic Induction

154743 Find the equivalent inductance between $A$ and $B$ in the given circuit.

1

1 H

2

4 H

3

0.8 H

4

16 H

Explanation:

A The diagram can be resolved as -

The combination is parallel -
Hence, $\frac{1}{L_{eq}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} + \frac{1}{L_{3}} + \frac{1}{L_{4}}$
$\frac{1}{L_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4}$
$L_{eq} = 1 H$

AP EAMCET-03.09.2021

Electro Magnetic Induction

154744 Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 H$ , and when connected in series are equivalent to a single inductor of inductance $8 H$ . Find the difference in the inductances of $A$ and $B$ .

1

3 H

2

7.5 H

3

2 H

4

4 H

Explanation:

D Given,
$L_{1} + L_{2} = 8 H \dots . (i) (connected in series)$
$\frac{1}{L_{1}} + \frac{1}{L_{2}} = \frac{1}{1.5} \dots . (ii) (connected in parallel)$
Also, $\frac{L_{1} L_{2}}{L_{1} + L_{2}} = 1.5$
$\frac{L_{1} (8 - L_{1})}{L_{1} + (8 - L_{1})} = 1.5$
$8 L_{1} - L_{1}^{2} = 12$
$L_{1}^{2} - 8 L_{1} + 12 = 0$
$L_{1}^{2} - 6 L_{1} - 2 L_{1} + 12 = 0$
$L_{1} (L_{1} - 6) - 2 (L_{1} - 6) = 0$
$L = 2, 6$
The difference between inductances $A$ and $B$ is $= 6 - 2 = 4 H$

AP EAMCET-19.08.2021