A. Suppose after charging a 11.4 Suppose after charging a 11.4 µF capacitor to 6

Question

A. Suppose after charging a 11.4

Suppose after charging a 11.4 µF capacitor to 6 V, you connect it to another already charged capacitor of 5.8 µF and 7 V. What voltage will your voltmeter read across those two capacitors connected in parallel? Express the answer with one decimal place. Five capacitors are connected across a potential difference Vab as shown in Fig. Because of the dielectrics used, each capacitor will break down if the potential across it is V=58 V. What should be the value of Vab (with one decimal place) if you reach the break down voltage one of the capacitor. Suppose after charging a 2.2 µF capacitor to 9 V, you connect it to the second uncharged capacitor of 9 µF. What voltage will your voltmeter read across the second capacitor after repeating the charge sharing process two times? Express the answer with one decimal place.

Explanation / Answer

You must use 4 principles in answering this: (a) since the caps are joined in parallel their final voltage (V) will be equal; (b) charge (Q) is conserved, so after having made the connection, the final charge wil be equal to the sum of the 2 initial charges; (c) Q = V*C in all cases (C = capacitance); (d) when caps are joined in parallel the total capacitance (Ctot) will be the sum of the individual capacitances.

C1 = 11.4*6 = 68.4uC C2 = 5.8*7 = 40.6uC. Total charge = 109uC

V = 109/(11.4 + 5.8) = 6.3V

c)

Equation for potential equivalent across two capacitance:

V = (C1V1 + C2V2) / (C1 + C2)

We first need to find V2, volts across second capacitance:
C1 = 3.7, V1 = 6V, C2 = 12
Unconnected, V2 = 0
V = (2.3 * 9 + 9 * 0) / (2.2 + 9) = 1.814 volts
Take this charge and plug it in for V2

V = (2.2 * 9 + 9 * 1.814) / (2.2 + 9) = 3.225 V

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