A) You have two capacitors, one with capacitance 17.7 %u03BCF and the other of u

Question

A)

You have two capacitors, one with capacitance 17.7 %u03BCF and the other of unknown capacitance. You connect the two capacitors in series with a voltage difference of 327-V applied across the capacitors. You discover that, as a result, the unknown capacitor has a charge of 1.21 mC. Find its capacitance.

B)

A parallel combination of a 1.13-%u03BCF capacitor and a 2.51-%u03BCF one is connected in series to a 4.35-%u03BCF capacitor. This three-capacitor combination is connected to a 19.3-V battery. Find the charge of each capacitor.

Explanation / Answer

A) The charge will be same on both capacitors...

SO, the charge on 17.7 uF capacitor is also 1.21 mC

So, voltage across it :

V = Q / C = 68.36 V

So the voltage across unknown capacitor = 327 - 68.36 = 258.64 Volts

Q = 1.21 mC

So, C = Q / V = 4.678 *10^-6 F = 4.678 uF

B) 1.13 || 2.51

So, C eq = 1.13 + 2.51 = 3.64 uF

It is in series with 4.35 uF

So C_eq = 3.64 * 4.35 / (3.64 + 4.35) = 1.977 uF

V = 19.3 V

So, Q total = C*V = 38.16 uC

So, Voltage drop across 4.35 uF = Q / C = 38.16 / 4.35 = 8.77 Volts

So, voltage drop across the combination of 1.13 || 2.51 = 19.3 - 8.77 = 10.52 Volts

So, charge on 1.13 uF capacitor = C*V = 1.13 * 10.52 = 11.89 uC

So, charge on 2.51 uC capacitor = C*V = 2.51 * 10.52 = 26.27 uC

SO, answers :

1.13 uC---------->11.89 uC charge

2.51 uC---------->26.27 uC charge

4.35 uC---------->38.16 uC charge

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