In Fig. 25-31 a15 V battery is connected across capacitors of capacitances C1 = C6 = 3.0 μF and C3 = C5 = 2.0C2 = 2.0C4 = 6.0 μF. What are the(a) potential V3 across and(b) charge q3 (in C) on capacitor 3?
In order to find V3 and Q3, you need to find the voltage over C4. Since C4 is in parallel with the caps C3 and C5, the voltage over them will be the same. To find V4, you need to figure out the total capacitance of C4 and C6, since the voltage over those two in series will be 15V, since the series combination is in parallel with the voltage source. C46 = 1/[(1/C4) + (1/C6)] = 2uF Q46 = C46V = 2uF*15V = 30uC = Q4 (charge over capacitors in series is the same) V4 = Q4/C4 = 30uC/6uF = 5V Now you know that there are 5V over the series combination of C3 and C5. Since C3 and C5 are the same value capacitor, the voltage over them is split equally and thus V3 = 5V/2 = 2.5V. Q3 = C3V3 = 2uF*2. 5V = 5uC = 5x10-6C
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Combine different combinations ofCapacitors to figure out the total equivalent capacitance of the circuit. C35 = 1/[(1/2)+(1/2)] = 1uF C235 = 1 + 2 = 3uF C2345 = 3 + 6 = 9uF C16 = 3 + 3 = 6uF C123456 = 1/[(1/9)+(1/6)] = 3.6uF total Q of the circuit = CV = 3. 6*15 = 54uC Charge Q over the capacitor combinations C2345 and C16 is going to be 54uC over each combo, since they are in series. V2345 = Q2345/C2345 = 54/9 = 6V = V4 = V35 = V2 (voltage over the capacitors C4, C35, and C2 are all going to be 6V since the combos are all in parallel) Since C3 and C5 are equal, they split the voltage equally and so V3 = 3V
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