question_answer

The charge on capacitors shown in figure and the potential difference across each will be respectively

A)  \[\text{24}0\mu \text{C},\text{8}0\mu \text{C},\text{16}0\mu \text{C and 8}0\text{ V},\text{ 4}0\text{ V},\text{ 4}0\text{V}\]

B)  \[\text{3}00\mu \text{C},\text{75}\mu .\text{C},\text{15}0\mu \text{C and 4}0\text{ V},\text{ 8}0\text{ V},\text{ 6}0\text{V}\]

C)  \[\text{22}0\mu \text{C},\text{7}0\mu \text{C},\text{14}0\mu \text{C and 6}0\text{ V},\text{ 5}0\text{ V},\text{ 4}0\text{V}\]

D)  None of the above

Correct Answer: A

Solution :

                 In the given circuit, \[{{C}_{2}}\] and \[{{C}_{3}}\] are in parallel, hence equivalent capacitance is                 \[C{{C}_{2}}+{{C}_{3}}=2+4=6\,\mu F\]                 The equivalent capacitance                 \[\frac{1}{C}=\frac{1}{C}+\frac{1}{{{C}_{1}}}\] \[\Rightarrow \]               \[C=\frac{{{C}_{1}}C}{{{C}_{1}}+C}\]                 \[=\frac{3\times 6}{3+6}=2\mu F\] Charge on combination,                 \[q=CV=2\times 120=240\,\mu F\] Charge on           \[{{C}_{1}}=240\,\mu C\] PD across \[{{C}_{1}}\] is,                 \[{{V}_{1}}=\frac{f}{{{C}_{1}}}=\frac{240\mu C}{3\mu F}=80V\] PD across C is                                 \[{{V}_{2}}=\frac{q}{C}=\frac{240\,\mu C}{6\mu \,C}=40\,V\]                     Charge on \[{{C}_{2}}\] is                 \[{{g}_{2}}={{C}_{2}}{{V}_{2}}\]                 \[=2\mu F\times 40=80\,\mu C\] Charge on \[{{C}_{3}}\] is                 \[{{Q}_{3}}={{C}_{3}}{{V}_{3}}\]                 \[=4\mu F\times 40=160\,\mu C\]. Hence, charges on \[{{C}_{1}},{{C}_{2}},{{C}_{3}}\] are \[240\,\mu C\], \[80\mu C,\,\,\,160\,\mu C\] and PD across \[{{C}_{1}},{{C}_{2}},{{C}_{3}}\] are 80 V, 40 V, 40 V respectively.