Two capacitors are connected as shown in the figure,with C₁ = 4.0 µF and C₂ = 7.0 µF.If a voltage source V = 90 V is applied across the combination,find the potential difference across C₁. A)57 V B)36 V C)60 V D)9.0 V

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Quizplus · Accepted Answer

The total capacitance of the combination can be found by adding the capacitors in series: 1/Ctot = 1/C1 + 1/C2 1/Ctot = 1/4.0 µF + 1/7.0 µF Ctot = 2.55 µF We can then use the equation for the potential difference across a capacitor: V = Q/C where V is the potential difference, Q is the charge on the capacitor, and C is the capacitance. Since the capacitors are in series, they have the same charge on them, so we can write: Q = Ctot * V For C₁: V1 = Q/C1 = (Ctot * V)/(4.0 µF) V1 = (2.55 µF * 90 V)/(4.0 µF) V1 = 57 V Therefore, the potential difference across C₁ is 57 V.

Two Capacitors Are Connected as Shown in the Figure,with C1

Two Capacitors Are Connected as Shown in the Figure,with C₁