- What is the initial battery current
*immediately after*switch*S*is closed? - What is the battery current a
*long time after*the switch is closed? - After the switch is re-opened, how long does it take for the current in the 600 ohm resistor to fall to 1.0 mA?

- At
*t*= 0, the capacitor is uncharged, which means that its voltage is zero. Accordingly, the voltage across the 600 ohm resistor is also zero, since they are in parallel. The uncharged capacitor effectively shorts out that resistor, leaving the emf of 36 V across the 1.2 k resistor. Thus the battery current is (36 V)/(1.2 k) = 30 mA initially.(This is the current that starts to charge the capacitor; as the voltage on the capacitor builds, the current through the 600 ohm resistor will grow accordingly.)

- When the capacitor is fully charged, no more current will flow into the capacitor branch from
the junction. Thus it is like an open circuit and the effective circuit becomes the two
resistors in series with the battery -- when the capacitor has reached its final charge state.
Now the battery current is
(36 V)/(1.8 k) = 20 mA.
- The circuit becomes the simple RC circuit for a charged capacitor discharging through
a resistor. The time constant is RC = (600 ohm)(250 x 10
^{-6}F) = 150 ms.The current decays as i(t) = i

_{0}exp(-t/RC). The initial current is 20 mA from part b. We seek the time that it takes for i(t)/i_{0}to reach 1.0 mA. Solving algebraically, that is the time RC ln(20) = 450 ms.

"http://www.physics.udel.edu/~watson/phys208/quiz5soln.html"

Last updated Oct. 18, 1997.

Copyright George Watson, Univ. of Delaware, 1997.