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In the figure below, a potential difference V = 200 V is applied across a capaci

ID: 1534611 • Letter: I

Question

In the figure below, a potential difference V = 200 V is applied across a capacitor arrangement with capacitances C1 = 11.0 F, C2 = 7.00 F, and C3 = 3.00 F. Find the following values.

(a) the charge on capacitor 3
C

(b) the potential difference across capacitor 3
V

(c) the stored energy for capacitor 3
J

(d) the charge on capacitor 1
C

(e) the potential difference across capacitor 1
V

(f) the stored energy for capacitor 1
J

(g) the charge on capacitor 2
C

(h) the potential difference across capacitor 2
V

(i) the stored energy for capacitor 2

o

Explanation / Answer


Given data

V = 200 v, C1 = 11.0*10^-6 F, C2 = 7.0*10^-6 F, C3 = 3.0*10^-6 F

(a) charge on the capacitor C3

Q3 = C3 * V

= 3*10^-6 * 200

= 600*10^-6 F

(b) potential difference across capacitor C3

V3 = V

= 200 volt

(c) energy stored in the capacitor E3

E3 = 0.5 * C3 * V3^2

= 0.5 * 3.0*10^-6 * 200^2

= 0.060 J

(d) charge on C1 = charge on C2

C1*V1 = C2*V2

V1/V2 = C2/C1 = 7.0/11.0

Hence, V1 = (7/11)*200

= 127.27 v

V2 = 200 - 127.27

= 72.73 v

charge on C1 = C1*V1

= 11*10^-6 * 127.27

= 1399.97*10^-6 F

(e) potential difference across C1

V1 = 127.27 V

(f) energy stored in the capacitor C1

E1 = 0.5 * C1 * V1^2

= 0.5 * 11.0*10^-6 * 127.27^2

= 0.089 J

(g) charge on C2

= C2 * V2

= 7*10^-6 * 72.73

= 509.11*10^-6 F

(h) potential difference across C2

V2 = 72.73 v

(i) energy stored in the capacitor C2

E2 = 0.5 * C2 * V2^2

= 0.5 * 7*10^-6 *72.73^2

= 0.0185 J