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.You need to login to perform this action.
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