| A network of four capacitors of capacity equal to \[{{C}_{1}}=C,\text{ }{{C}_{2}}=2C,\text{ }{{C}_{3}}=3C\] and \[{{C}_{4}}=4C\] are connected to a battery as shown in the figure The ratio of the charges on \[{{C}_{2}}\] an \[{{C}_{4}}\] is: [AIPMT (S) 2005] |
 |
A) \[\frac{22}{3}\]
B) \[\frac{3}{22}\]
C) \[\frac{7}{4}\]
D) \[\frac{4}{7}\]
Correct Answer: B
Solution :
| [b] Key Idea: Charge on a capacitor is the product of capacitance and potential difference across it. |
| The charge flowing through \[{{C}_{4}}\] is |
| \[{{q}_{4}}={{C}_{v}}\times V=4CV\] |
| The series combination of \[{{C}_{1}},\text{ }{{C}_{2}}\] and \[{{C}_{3}}\] gives |
| \[\frac{1}{C'}=\frac{1}{C}+\frac{1}{2C}+\frac{1}{3C}\] |
| \[=\frac{6+3+2}{6C}=\frac{11}{6C}\] |
| \[\Rightarrow \] \[C'=\frac{6C}{11}\] |
| Now, \[C'\] and \[{{C}_{4}}\] form parallel combination giving |
| \[C''=C'+{{C}_{4}}\] |
| \[=\frac{6C}{11}+4C=\frac{50\,C}{11}\] |
| Net charge \[q=C''V\] |
| \[=\frac{50}{11}CV\] |
| Total charge flowing through \[{{C}_{1}},\text{ }{{C}_{2}},\text{ }{{C}_{3}}\] will be |
| \[q'=q-{{q}_{4}}\] |
| \[=\frac{50}{11}CV-4C\,V=\frac{6CV}{11}\] |
| Since, \[{{C}_{1}},\text{ }{{C}_{2}}\] and \[{{C}_{3}}\] are in series combination hence, charge flowing through these will be same. |
| Hence, |
| \[{{q}_{2}}={{q}_{1}}={{q}_{3}}=q'=\frac{6\,C\,V}{11}\] |
| Thus, \[\frac{{{q}_{2}}}{{{q}_{4}}}=\frac{6CV/11}{4CV}=\frac{3}{22}\] |
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