Question

In the circuit shown in the figure, the battery is ideal, = 12.6 V, L = 11.3 mH, R1 = 11.1 ?, and R2 = 20.1 ?. The switch has been open for a long time before it is closed at t = 0. At what rate is the current in the inductor changing (a) immediately after the switch is closed and (b) when the current in the battery is 0.530 A? (c) What is the current (in A) in the battery when the circuit reaches its steady-state condition?

Answer 1

a)

Immediately after switch is closed ,inductor will act as open circuit ,so Voltage across the Resistor R1 is

V₁=E*(R₁/R₁+R₂) =12.6*(11.1/11.1+20.1)

V₁ = 4.48 Volts

Voltage across inductor is

V_L=V₁=4.48

=>L(dI/dt) =4.48

dI/dt =4.48/(11.3*10^-3)

dI/dt =396.7 A/s

b)

Voltage across R₁ is

V₁=IR₁ =0.53*11.1 =5.883 Volts

Rate of Change in current

dI/dt =5.883/(11.3*10^-3)

dI/dt= 520.6 A/s

c)

After long time ,inductor will act as short circuit ,so current in the battery is

I=E/R₂=12.6/20.1

I=0.627 A