Put the capacitors in series with the 4.5 V of batteries in series. Although no DC current will flow, the voltage drop across the two capacitors will be in inverse proportion to the capacitances.
C1/C2 = V2/V1
With one known capacitance C1, you can solve for C2.
You do not need the light bulb, resistor or ammeter.
How do you measure a unidentified capacitor when you have a identified one as well as voltmeter, a ammeter, a resistor and a light bulb as well as 3 1,5 batterycells. You have a conductinhcords?
1 answer
1 answer