In the figure R 1 = 10.8 k?, R 2 = 15.6 k?, C = 0.428 ?F, and the ideal battery

Question

In the figure R1 = 10.8 k?, R2 = 15.6 k?, C = 0.428 ?F, and the ideal battery has emf ? = 19.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

In the figure R1 = 10.8 k?, R2 = 15.6 k?, C = 0.428 ?F, and the ideal battery has emf ? = 19.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 4.00 ms?

Explanation / Answer

First calculate the initial current in R2 at the time the switch is opened as Io = E/(R1 + R2) = 19/(26400) = 719.69 uA

then calculate the R*C time constant as tau = R2*C. tau = 15600*0.428*10^-6 = 6.67 msec.

Then use I(t) = Io*exp(-t/tau) with t = 4ms. This gives I = 719.69*exp(-4/6.67) = 395.334 uA

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