Question

# Two condensers $${C}_{1}$$ and $${C}_{2}$$ in a circuit are joined as shown in the figure. The potential of point $$A$$ is $${V}_{1}$$ and that of $$B$$ is $${V}_{2}$$. The potential of point $$D$$ will be :

A
12(V1+V2)
B
C1V2+C2V1C1+C2
C
C1V1+C2V2C1+C2
D
C2V1+C1V2C1+C2

Solution

## The correct option is D $$\cfrac { { C }_{ 1 }{ V }_{ 1 }+C_{ 2 }{ V }_{ 2 } }{ { C }_{ 1 }+C_{ 2 } }$$$${V}_{1}-{V}_{2}$$ will be divided between $${C}_{1}$$ and $${C}_{2}$$ and $$V_D$$ is potential at point D in series. Potential difference across $${C}_{1}$$ is$${ V }_{ 1 }-{ V }_{ D }=\cfrac { { C }_{ 2 }({ V }_{ 1 }-{ V }_{ 2 }) }{ { C }_{ 1 }+{ C }_{ 2 } }$$$${V }_{ D }={ V }_{ 1 }-\cfrac { { C }_{ 2 }({ V }_{ 1 }-{ V }_{ 2 }) }{ { C }_{ 1 }+{ C }_{ 2 } }$$or $${ V }_{ D }=\cfrac { { C }_{ 1 }{ V }_{ 1 }+{ C }_{ 2 }{ V }_{ 2 } }{ { C }_{ 1 }+{ C }_{ 2 } }$$Physics

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