In the figure below, a potential difference V = 100 V is applied across a capaci

ID: 2018266 • Letter: I

Question

In the figure below, a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 10.0 µF, C2 = 6.00 µF, and C3 = 4.00 µF. Find the following values.

(a) the charge on capacitor 3

(b) the potential difference across capacitor 3

(c) the stored energy for capacitor 3

(d) the charge on capacitor 1

(e) the potential difference across capacitor 1

(f) the stored energy for capacitor 1

(g) the charge on capacitor 2

(h) the potential difference across capacitor 2

(i) the stored energy for capacitor 2

Explanation / Answer

Data: C1 = 10 x 10^-6 F C2 = 6 x 10^-6 F C3 = 4 x 10^-6 F Voltage, V = 100 V Solution: Charge of C1 = charge of C2 Q1 = Q2 C1 V1 = C2 V2 V1 / V2 = C2 / C1 V1 / V2 = 0.6 V1 = 0.6 V2 But, V1 + V2 = V 0.6 V2 + V2 = V V2 = V / 1.6 = 100 / 1.6 = 62.5 V So, V1 = 100 - 62.5 = 37.5 V V3 = V = 100 V (a) Q3 = C3 V = 4 x 10^-6 * 100 = 400 x 10^-6 C (b) V3 = V = 100 V (c) Energy stored of C3 = (1/2) C3 V3^2 = 0.5 * 400 x 10^-6 * 100^2 = 200 x 10^-2 = 2 J (d) Q1 = C1 V1 = 10 x 10^-6 * 37.5 = 375 x 10^-6 C (e) V1 = 37.5 V (f) Energy stored of C1 = (1/2) C1 V1^2 = 0.5* 10 x 10^-6 * 37.5^2 = 0.007 J (g) Q2 = C2 V2 = 6 x 10^-6 * 62.5 = 35 x 10^-6 C (h) V2 = 62.5 V (i) Energy for C2 = (1/2) C2 V2^2 = 0.5 * 6 x 10^-6 * 62.5^2 = 0.012 J

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