A 10-V-emf battery is connected in series with the following: a2−μF capacitor, a 2-ohm resistor, an ammeter, and a switch,initially open; a voltmeter is connected in parallel a cross the capacitor. At the instant the switch is closed, what are the current and capacitor voltage readings, respectively?

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Margot Mill · Accepted Answer

Given that  The emf of the battery (V) = 10V  Resistance in the circuit is given by (R) = 2?  At t=0 capacitor acts like a short circuit  Therefore we get the current passing through the circuit is given by  I=V/R  =10/2  =5 A  Then the Capacitor voltage Vc=0

A 10-V-emf battery is connected in series with the following: a2-\mu F capacitor, a 2-ohm resistor, an ammeter, and a switch,initially open; a voltmet

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