Question

# A capacitor of capacitance $${C}_{1}=1\mu F$$ charged upto a voltage $$V=110V$$ is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances $${C}_{2}=2\mu F$$ and $${C}_{3}=3\mu F$$. Then, the amount of charge that will flow through the connecting wires is :

A
40μC
B
50μC
C
60μC
D
110μC

Solution

## The correct option is D $$60\mu C$$Initial charge on $${C}_{1}$$ is $${Q}_{1}={C}_{1}V=110\mu C$$Let $$x$$ charge flow through wires.$$\dfrac { { Q }_{ 1 }-x }{ { C }_{ 1 } } =\dfrac { x }{ { C }_{ eq } }$$$${ C }_{ eq }=\dfrac { { C }_{ 2 }\times { C }_{ 3 } }{ { C }_{ 2 }+{ C }_{ 3 } } =\dfrac{{2}\times {3}}{2+3}=\dfrac{6}{5}\mu F$$$$\dfrac { { Q }_{ 1 }-x }{ { C }_{ 1 } } =\dfrac { x }{ { C }_{ eq } }$$$$\dfrac{110-x}{1}=\dfrac{5x}{6}$$$$\Rightarrow x=60\ \mu C$$Physics

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