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Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

In the circuit shown in figure (1) the charge on\[{{C}_{2}}\]is greater than on \[{{C}_{1}}\] (2) the charges on\[{{C}_{1}}\]and\[{{C}_{2}}\]are same (3) potential difference across\[{{C}_{1}}\]and\[{{C}_{2}}\]are same (4) potential difference across\[{{C}_{1}}\]is greater than across\[{{C}_{2}}\]

A)  1, 2 and 3 are correct

B)  1 and 2 are correct

C)   2 and 4 are correct

D)  1 and 3 are correct

Correct Answer: C

Solution :

                 Since there is only one path for the charge flow, capacitors\[{{C}_{1}}\]and\[{{C}_{2}}\]are in series via the cells of emf.\[{{E}_{1}}\]and \[{{E}_{2}}\]                                 The net emf in the circuit                 \[E={{E}_{1}}-{{E}_{2}}=12-6=6V\]                 The equivalent capacity                 \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]                 \[=\frac{4\times 8}{4+8}=\frac{8}{3}\mu F\]                 Charge on each capacitor\[(q={{q}_{1}}={{q}_{2}})\]                 \[q=CV=\frac{{{C}_{1}}{{C}_{2}}({{E}_{1}}-{{E}_{2}})}{{{C}_{1}}+{{C}_{2}}}=\frac{8}{3}\times 6=16\,\mu F\]                 For a capacitor\[q=CV\]or \[V=\frac{q}{C}\]                 \[\therefore \]\[{{V}_{1}}=\frac{{{q}_{1}}}{{{C}_{1}}}=\frac{q}{{{C}_{1}}}=\frac{{{C}_{2}}({{E}_{1}}-{{E}_{2}})}{{{C}_{1}}+{{C}_{2}}}=\frac{16}{4}=4\,V\]                 \[{{V}_{2}}=\frac{{{q}_{2}}}{{{C}_{2}}}=\frac{q}{{{C}_{2}}}=\frac{{{C}_{1}}({{E}_{1}}-{{E}_{2}})}{{{C}_{1}}+{{C}_{2}}}=\frac{16}{8}=2\,V\]