A 12 V battery is connected to two resistors (R_1= 5 ohms R_2= 7 ohms) and an in

Question

A 12 V battery is connected to two resistors (R_1= 5 ohms R_2= 7 ohms) and an inductor ( L= 3mH). Switch S has been open for a long time and is closed at t=0 s. (a) What is the magnitude of the current I_L immediately after the switch is closed (b) What is the magnitude of the current dI_L/dt immediately after the switch is closed (c)What is the magnitude of the current I_L after the switch is closed for a long time (d) After the switch has been closed for a long time, it is opened. The current I_L now decays exponentially as a function of time. What is the value of the time constant of this decay?

Explanation / Answer

a)

Immediately afer switch is closed ,inductor acts as open circuit ,so current through inductor is

IL=0 A

b)

Total Current flowing in the circuit

I=V/(R1+R2) =12/(5+7) =1 A

Voltage across Inductor

VL=V1=I*R1=1*7

VL=7 Volts

since

VL=LdIL/dt

=>dIL/dt =VL/L =7/(3*10-3) =2333.33A/s

c)

After Long time indcutor acts as short circuit ,so current flowing through inductor is

IL=V/R1=12/5 =2.4 A

d)

New time Constant

T=L/R2=3*10-3/7

T=4.2857*10-4 seconds or 0.42857 ms

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