Question

Consider the following figure. A circuit consists of three square loops with a total of seven capacitors. The circuit begins with wire going down from a point a. The wire splits three ways forming two identical square loops with a shared side before reconnecting. The loops each contain a capacitor C1 on the top side, a capacitor C2 on the bottom side, and a capacitor C3 on the shared side. After reconnecting, the wire splits again into another square loop. The loop contains two capacitors labeled C2, one on the left side and one on the right side. The wire reconnects again and ends at a point b.(a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in the figure above. Take C1 = 7.00 µF, C2 = 15.0 µF, and C3 = 1.00 µF. µF (b) What charge is stored on C3 if the potential difference between points a and b is 60.0 V? µC

Answer 1

The equivalent capacitance between points a and b is 53.50 µF, and the charge on capacitor C3 is 60.0 µC.

(a) Equivalent Capacitance

To find the equivalent capacitance between points a and b, we can break down the circuit into smaller sections and use the rules for combining capacitors in series and parallel.

First, consider the two identical square loops on the left. The capacitors in each loop are in parallel, so the equivalent capacitance of each loop is:

C_loop = C1 + C2 + C3 = 7.00 µF + 15.0 µF + 1.00 µF = 23.00 µF

Since the two loops are identical, their equivalent capacitances are equal. Therefore, the combined equivalent capacitance of these two loops is simply twice the equivalent capacitance of one loop:

C_combined = 2 * C_loop = 2 * 23.00 µF = 46.00 µF

Next, consider the square loop on the right. The two capacitors in this loop are in series, so their equivalent capacitance is:

C_right = 1 / (1/C2 + 1/C2) = 1 / (2/15.0 µF) = 7.50 µF

Finally, the equivalent capacitance between points a and b is the combination of C_combined and C_right in parallel. Therefore:

C_total = C_combined + C_right = 46.00 µF + 7.50 µF = 53.50 µF

(b) Charge on C3

The charge on capacitor C3 can be found using the formula:

Q = CV

where:

Q is the charge (C)

C is the capacitance (F)

V is the potential difference (V)

In this case, the capacitance of capacitor C3 is 1.00 µF, and the potential difference between points a and b is 60.0 V. Therefore, the charge on capacitor C3 is:

Q_C3 = C3 * V = 1.00 µF * 60.0 V = 60.0 µC

Therefore, the equivalent capacitance between points a and b is 53.50 µF, and the charge on capacitor C3 is 60.0 µC.