A coil having a resistance of 20 ohm and inductance 20 henry is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached

Question

A coil having a resistance of 20 ohm and inductance 20 henry is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached (A) 250 J (B) 150 J (C) 300 J (D) 360 J

Tardigrade Admin · Accepted Answer

Here, R = 20Ω, L = 20 H and E = 120 V ∴ I0 = (E/R) = (120/20) = 6A Energy stored = (1/2) LI0 2 = (1/2) × 20 × 36 = 360 J

Q. A coil having a resistance of 20 ohm and inductance 20 henry is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached

250 J

150 J

300 J

360 J

Here, R = 20 $Ω$ , L = 20 H and E = 120 V
$∴$ $I_{0} = \frac{E}{R} = \frac{120}{20}$ = 6A
Energy stored = $\frac{1}{2} L I_{0}^{2} = \frac{1}{2} \times 20 \times 36$ = 360 J

Q. A coil having a resistance of 20 ohm and inductance 20 henry is connected to 120 V battery. Find the value of energy stored in the magnetic field when the steady state is reached

250 J

150 J

300 J

360 J

Here, R = 20Ω, L = 20 H and E = 120 V ∴ I0​=RE​=20120​ = 6A Energy stored = 21​LI0​2=21​×20×36 = 360 J

Here, R = 20 $Ω$ , L = 20 H and E = 120 V
$∴$ $I_{0} = \frac{E}{R} = \frac{120}{20}$ = 6A
Energy stored = $\frac{1}{2} L I_{0}^{2} = \frac{1}{2} \times 20 \times 36$ = 360 J