• # question_answer A constant potential difference between A and B is applied. When the switch is shifted from position P to position Q, then A)  charge on $2\mu F$ capacitor is approximately$13\mu C$ B)  potential difference across $3\mu C$capacitor is approximately 4.3 V  C)  potential difference across $1\mu F$ capacitor is approximately 11 V D)  All of the above

Solution :

Initially,           ${{Q}_{1}}=24\mu \,C$ Finally, ${{V}_{C}}=\frac{24}{\frac{6}{5}+1}=\frac{24\times 5}{11}\,volt$ $Q_{3}^{'}=Q_{2}^{'}=\frac{6}{5}\times \frac{24\times 5}{11}13\mu \,C$  (approx) $Q_{1}^{'}=1\times \frac{24\times 5}{11}=1\,\,1\mu \,C$                      (approx) $V_{1}^{'}=11$ volt $V_{3}^{'}=\frac{Q_{3}^{'}}{{{C}_{3}}}=\frac{13}{3}=4.3\,$ volt                       (approx)

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