Question
Consider the circuit shown with key opened. 3uf capacitor is charged to p.d. of 9 V. Capacitor 6 uF is charged to p.d. of 3V. The key is now closed . What will be the potential differnce across 3 uF capacitor in steady state?
Joshi sir comment
V = [3*9-6*3]/(3+6) = 9/9 = 1
here - is used for opposite poles of the capacitors are connected together
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GAURAV MAHATE 12 year ago
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