Question

The circuit in the figure below contains a 9.00 v battery and four capacitors. The two capacitors on the left and right both have same capacitance of C, -14.20. The capacitors in the top two branches have capacitances of 6.00 pF and C - 28.20 F 6.00 с. 9.00 V (a) What is the equivalent capacitance (in pf) of all the capacitors in the entire circuit? (b) What is the charge (in) stored by each capacitor? right 34.20 pf capacitor uc left 34.20 pf capacitor C 28.20 pF capacitor 6,00 capacitor C (c) What is the potential difference in V) across each capacitor? (Enter the magnitudes.) right 34.20 pF capacitor V left 34.20 uF capacitor V 28.20 pF capacitor 6.00 uF capacitor V

Answer 1

6C + + (a) The equivalent capacitance is 2C Cea 6+C 2+C 6(34.20), 2(28.20) 6+34.20 2+28.20 = 6.97 uF (b) The charges are Qup 6(34.20) (9.0) 6+34.20 = 45.94 uc Lion = Cion V 2(28.20) (9.0) 2+28.20 = 16.81 °C = 44.94 uc Q2 = 16.81uc Qouc = 44.94 uc Q2uc = 16.81°C (c) The voltages are V = &_44.94 C 34.20 = 1.31 V 16.81 Q2 V2 C2 28.20 = 0.597 V Vz Q_34.20 6 6 = 5.7 V 34,20 = V= 2 16.81 2 = 2.80 V