Question

A 10-V-emf battery is connected in series with the following: a 2-µF capacitor, a 2.0-Ω resistor, an ammeter, and a switch, initially open; a voltmeter is connected in parallel across the capacitor. At the instant the switch is closed, what are the current and capacitor voltage readings, respectively?

Answer 1

Answer:

I = 5.0 A  V=0

Explanation:

Assuming that the ammeter is an ideal one (which means that the internal resistance is negligible compared with other resistors in the circuit), as the voltage through the capacitor can't change instantaneously, just after the switch is closed, will behave like a short, so applying Ohm's law, the current in the circuit will be as follows:

$I =\frac{V}{R} = \frac{10 V}{2.0 \Omega} = 5.0 A$

As the voltage in the capacitor, can't change instantaneously, assuming an ideal voltmeter (infinite resistance) , the lecture on the voltmeter across the capacitor will be just 0.

As time goes by, the voltage measured will follow the following equation:

$V = V0*( 1-e^{-\frac{t}{R*C}} )$

We see that when t=0, V=0.