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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 25 of 37
Marks: +1, -0
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor.
Solution:  
Since C2C_2 and C3C_3 are in series, so, C' = 100 pF
Since C1C_1 and C' are parallel
so C" = C1C_1 + C' = 100 + 100 or C" = 200 pF
Since C4C_4 and C" are in series, so net capacitance of the network is
1C\frac{1}{C} = 1C′+1C4\frac{1}{C'} + \frac{1}{C_4} = 1200+1100\frac{1}{200} + \frac{1}{100} = 1+2200\frac{1+2}{200} or C2C_2 = 2003\frac{200}{3} pF = 66.7 pF
Net charge stored on the combination is
Q = CV = 2003\frac{200}{3} × 10−1210^{-12} × 300 = 2 × 10−810^{-8} C
As C″ and C4C_4 are in series, so Q" = Q4Q_4 = Q or Q" = Q4Q_4 = 2 × 10−810^{-8} C
and hence V" = Q′C′\frac{Q'}{C'} = 2×10−8 C200×10−12 F\frac{2 \times 10^{-8}\,\mathrm{C}}{200 \times 10^{-12}\,\mathrm{F}} = 100 V
and V4V_4 = Q4C4\frac{Q_4}{C_4} = 2×10−8 C100×10−12 F\frac{2 \times 10^{-8}\,\mathrm{C}}{100 \times 10^{-12}\,\mathrm{F}} = 200 V
Since C1C_1 and C' are in parallel, so
V1V_1 = V' = V" or V1V_1 = V' = 100 V
and hence Q1Q_1 = C1V1C_1 V_1 = 100 × 10−1210^{-12} × 100 = 1 × 10−810^{-8} C
and Q' = C1V1C_1 V_1 = 100 × 10−1210^{-12} × 100 = 1 × 10−810^{-8} C
and Q' = C'V' = 100 × 10−1210^{-12} × 100 = 1 × 10−810^{-8} C
Since C2C_2 and C3C_3 are in parallel, so Q2Q_2 = Q3Q_3 = Q' or Q2Q_2 = Q3Q_3 = 1 × 10−810^{-8} C and hence V2V_2 = Q2C2\frac{Q_2}{C_2} = 1×10−8 C200×10−12 F\frac{1 \times 10^{-8}\,\mathrm{C}}{200 \times 10^{-12}\,\mathrm{F}} = 50 V
and V3V_3 = Q3C3\frac{Q_3}{C_3} = 1×10−8 C200×10−12 F\frac{1 \times 10^{-8}\,\mathrm{C}}{200 \times 10^{-12}\,\mathrm{F}} = 50 V
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