Question

The circuit in the figure below contains a 90.0 V battery and four capacitors. In the top parallel branch, there are two capacitors, one with a capacitance of C1 = 3.00 µF and another with a capacitance of 6.00 µF. In the bottom parallel branch, there are two more capacitors, one with a capacitance of 2.00 µF and another with a capacitance of C2 = 6.00 µF. A circuit consists of a 90.0 V battery and four capacitors. The wire begins at the positive terminal of the battery and splits into two parallel branches before reconnecting and then ending at the negative terminal of the battery. Each branch contains two capacitors in series. One branch contains a capacitor labeled C1 followed by a 6.00 µF capacitor. The other branch contains a 2.00 µF capacitor followed by a capacitor labeled C2. (a) What is the equivalent capacitance (in µF) of the entire circuit? Which capacitors are in series, and which are in parallel? Find the equivalent capacitance for the two capacitors in series across each branch first. Then find the equivalent capacitance of the circuit from the two parallel equivalent capacitances. µF (b) What is the charge (in µC) on each capacitor? on C1 µC on C2 µC on the 6.00 µF capacitor µC on the 2.00 µF capacitor µC (c) What is the potential difference (in V) across each capacitor? across C1 V across C2 V across the 6.00 µF capacitor V across the 2.00 µF capacitor V

Answer 1