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

Question

Quizplus · Accepted Answer

The total capacitance of the combination can be found using the formula 
$C_{total} = \frac{C_1 C_2}{C_1 + C_2}$
where $C_1$ and $C_2$ are the capacitances of the individual capacitors. 
Plugging in the values, we get 
$C_{total} = \frac{4 	imes 7}{4 + 7} = 2.545$ µF

The charge on each capacitor will be the same as they are in parallel. The charge on a capacitor is given by the formula 
$Q = CV$
where Q is the charge on the capacitor, C is the capacitance, and V is the potential difference across the capacitor.

The total charge on the combination is 
$Q = C_{total}V = 2.545 	imes 90 = 229.04$ µC

Now, using the formula Q = CV, we can find the potential difference across each capacitor. For the 4 µF capacitor, 
$V = \frac{Q}{C} = \frac{229.04}{4} = 57.26$ V

Therefore, the potential difference across the 4 µF capacitor is 57 V (rounded to the nearest whole number). Answer choice A is the correct answer.

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

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