The capacitors shown above have values of C1 = 6.16-microfarads; C2 = 4.25-micro

Question

The capacitors shown above have values of C1 = 6.16-microfarads; C2 = 4.25-microfarads; C3 = 6.43-microfarads; C4 = 3.65-microfarads; C5 = 2.78-microfarads; C6 = 4.52-microfarads; the emf of the battery is 22.5-volts. What is the voltage drop across the capacitor C2?

Explanation / Answer

c3 n c4 in series =>Ceq= 6.43*3.65/10.08 =2.32 this eq is parallel to C5 =>Ceq = 2.32 + 2.78 = 5.1 this is in parallel with C2 =>Ceq=5.1+4.25 = 9.35 C1 n C6 are in parallel =>Ceq = 6.16+4.52 = 10.68 voltage drop across the capacitor C2= voltage drop across the capacitor 9.35 => (10.68/(10.68+9.35)) * 22.5 =>11.9 V

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