Consider the circuit shown in the figure. The battery has emf epsilon = 25 volts
ID: 1568098 • Letter: C
Question
Consider the circuit shown in the figure. The battery has emf epsilon = 25 volts and negligible internal resistance. The inductance is L = 0.80 H and the resistances are R_1 = 12 ohm and R_2 = 9.0 ohm. Initially the switch S is open and no currents flow. Then the switch is closed. What is the current in the two resistors just after the switch is closed? What is the current through the resistors long after the switch has been closed? Show all the steps for credit and also complete the table below.Explanation / Answer
Potential drop across inductor V(L) = - L di/dt
Just before closing the switch, current through inductor is zero. Just after after closing the switch, there can't be any current through inductor, as it means there is sudden change in current or di/dt is infinity and that means potential drop across inductor is infinite, which is not possible.
With no current through inductor at t= 0, total current passes through two resistors.Hence resistor are like connected in series. Current through two resistors is same and equal to
V/(R1 +R2) = 25/21 = 1.19 Amp
Hence t=0 , I1 = I2 = 1.19 Amp
2) At t= infinity there is steady currect in circuit. Hence di/dt = 0 and potential drop across inductor = 0.
As Inductor is connected in parallel with R1, potential drop across R1 is also zero. So I1 = 0.
Current through R2 is same as that through inductor and equal to
V/R2 = 25/9 = 2.78 Amp = I2