Question

Consider the circuit shown at left. At 0.2 seconds after the switch is closed, What is the current in the circuit? What is the voltage across the inductor at this time? What is the energy stored in the inductor at this time?

Answer 1

(a) We have relation for after time t

$i=\frac{V(1-e^{\frac{-Rt}{L}})}{R}$

putting all values

$i=\frac{24(1-e^{\frac{-6\times 0.2}{2}})}{6}$

i=1.80 A ANS

(b)

Voltage across resistor V_R=IR

= 1.80$\times$6

= 10.8 V

Voltage across inductor V_L=(24-10.8) =13.2V ANS

(c)

Energy stored E= (LI²)/2

= (2$\times$1.8²) / 2

= 3.24 J ANS