A) 6/14 and 50V
B) 3/7 and 30V
C) 7/3 and 30V
D) None of these
Correct Answer: B
Solution :
Given, \[{{C}_{1}}=6\mu F,\] \[{{C}_{2}}=14\mu F\] \[{{V}_{1}}=100volt\] Charge on capacitor \[q={{C}_{1}}{{V}_{1}}=6\times 100=600\mu C\] Redistribution of charge takes place in the ratio of their capacitances. \[\therefore \] \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{6}{14}=\frac{3}{7}\] Charge on capacitor \[{{C}_{1}}\] after touching and then \[{{C}_{1}}\]removed, \[{{q}_{1}}=\frac{3\times 600}{3+7}=180\mu C\] \[\therefore \] Potential at \[6\mu F\] capacitor
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