An induction coil stores 32 J of magnetic energy and dissipates energy as heat at the rate of 320 W. A current of 4 A is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery.

Question

An induction coil stores 32 J of magnetic energy and dissipates energy as heat at the rate of 320 W. A current of 4 A is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(1/2) L I2=32 ∵ I=4 A ∴ (1/2) × L ×(4)2=32 ⇒ L=4 H Now, P=I2 R ∴ 320=(4)2 × R ⇒ R=(320/16)=20 Ω ∴ Time constant =(L/R)=(4/20)=(1/5) =0.2 s

Q. An induction coil stores $32 J$ of magnetic energy and dissipates energy as heat at the rate of $320 W$ . A current of $4 A$ is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery.

$\frac{1}{2} L I^{2} = 32$
$∵ I = 4 A$
$∴ \frac{1}{2} \times L \times (4)^{2} = 32$
$\Rightarrow L = 4 H$
Now, $P = I^{2} R$
$∴ 320 = (4)^{2} \times R$
$\Rightarrow R = \frac{320}{16} = 20 Ω$
$∴$ Time constant $= \frac{L}{R} = \frac{4}{20} = \frac{1}{5}$
$= 0.2 s$

Q. An induction coil stores 32J of magnetic energy and dissipates energy as heat at the rate of 320W. A current of 4A is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery.

21​LI2=32 ∵I=4A ∴21​×L×(4)2=32 ⇒L=4H Now, P=I2R ∴320=(4)2×R ⇒R=16320​=20Ω ∴ Time constant =RL​=204​=51​ =0.2s

Q. An induction coil stores $32 J$ of magnetic energy and dissipates energy as heat at the rate of $320 W$ . A current of $4 A$ is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery.

$\frac{1}{2} L I^{2} = 32$
$∵ I = 4 A$
$∴ \frac{1}{2} \times L \times (4)^{2} = 32$
$\Rightarrow L = 4 H$
Now, $P = I^{2} R$
$∴ 320 = (4)^{2} \times R$
$\Rightarrow R = \frac{320}{16} = 20 Ω$
$∴$ Time constant $= \frac{L}{R} = \frac{4}{20} = \frac{1}{5}$
$= 0.2 s$