Grade 11Electrostatics

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

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Guiness Srivathsan

8 Years agoGrade 11

1 Answer

Eshan

8 Years ago

When 1 is connected, capacitor C1 gets charged to potential 60V attaining a charge equal to $60\mu C$ . When 1 is diconnected and 2 is connected, the capacitor C1 starts to get discharged into an effective capacitance of $2\mu F$ (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 capacitances $C_1$ and $C_{eff}$ .
Thus $Q_1+Q_{eff}=60\mu C$

Hence $Q_{eff}=Q_2+Q_3=40\mu C$