In Fig. 24.23, each capacitor has C = 5.30 µF, and Vab = +25.0 V. Figure 24.23 (
ID: 1694546 • Letter: I
Question
In Fig. 24.23, each capacitor has C = 5.30 µF, and Vab = +25.0 V.Figure 24.23
(a) Calculate the charge on each capacitor.
Q1 = ? C Q2 = ? C
Q3 = ? C Q4 = ? C
(b) Calculate the potential difference across each capacitor.
V1 = ? V V2 = ? V
V3 = ? V V4 = ? V
(c) Calculate the potential difference between points a and d. ? V
In Fig. 24.23, each capacitor has C = 5.30 mu F, and Vab = +25.0 V. Figure 24.23 (a) Calculate the charge on each capacitor. Q1 = ? C Q2 = ? C Q3 = ? C Q4 = ? C (b) Calculate the potential difference across each capacitor. V1 = ? V V2 = ? V V3 = ? V V4 = ? V (c) Calculate the potential difference between points a and d. ? V
Explanation / Answer
IN the above network the equivalent capacitance C^1 of series connected capacitors C1 and C2 areC^1 = C1*C2/C1+C2
= 5.3*5.3*10^-12/10.6*10^-6
= 2.6 µF the equivalent capacitance of parallel combination of C^1 and C3 Cad = 2.6+5.3 = 7.9 µF The entire circuit now reduces to two capacitors C4 and Cad in parallel Equivalent capacitance of total network = Cad+C4 = 7.9+5.3 = 13.2 µF _________________________________________________________ 1) Total charge = q = C V = 13.2 *25 = 330 *10^-6 C charge on C4 = 330 *10^-6 C potential difference across = C4 = q/C4 = 330 *10^-6 C /5.3*10^-6 = 62.2 V __________________________________________________________