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GUYS PLEASE REPLY ASAP!!!!!!!!!!!WHAT IS THE ANSWER?

Guiness Srivathsan , 6 Years ago
Grade 11
anser 1 Answers
Eshan

Last Activity: 6 Years ago

When 1 is connected, capacitor C1 gets charged to potential 60V attaining a charge equal to60\mu C. When 1 is diconnected and 2 is connected, the capacitor C1 starts to get discharged into an effective capacitance of2\mu F(formed by series combination of C2 and C3).

From Kirchoff’s voltage law, the voltage across the two capacitors will be same. Thus
\dfrac{Q_1}{C_1}=\dfrac{Q_{eff}}{C_eff} \implies \dfrac{Q_1}{Q_{eff}}=\dfrac{C_1}{C_eff}=\dfrac{1}{2}

The initial charge attained by C1 has actually been divided into capacitancesC_1andC_{eff}.
ThusQ_1+Q_{eff}=60\mu C

HenceQ_{eff}=Q_2+Q_3=40\mu C

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