Capacitors

JEE Main level, single correct, medium

40 of 180

Question 40 of 180

Three capacitors C1=2μF,C2=6μF{C}_{1}=2\mu F,{C}_{2}=6\mu F and C3=12μF{C}_{3}=12\mu F are connected as shown in the figure. Find the ratio of the charges on capacitors C1,C2{C}_{1},{C}_{2} and C3{C}_{3} respectively.

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Answer: (D) 1:2:21:2:2

From the above circuit diagram, the equivalent capacitance of the combination of capacitors is,
Ceq=C1+C2C3C2C3{C}_{eq}={C}_{1}+\dfrac{{C}_{2}{C}_{3}}{{C}_{2}{C}_{3}},
Ceq=2+6×126+12=6μF{C}_{eq}=2+\dfrac{6\times 12}{6+12}=6\mu F
Then, QC1QC2=C1VC2C3C2+C3V=2×(6+12)6×12=12\dfrac{{Q}_{{C}_{1}}}{{Q}_{{C}_{2}}}=\dfrac{{C}_{1}V}{\dfrac{{C}_{2}{C}_{3}}{{C}_{2}+{C}_{3}}V}=\dfrac{2\times (6+12)}{6\times 12}=\dfrac{1}{2}
and QC2=QC3{Q}_{{C}_{2}}={Q}_{{C}_{3}}
So, the ratio of charges are, 1:2:21:2:2.

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