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Physics
In the circuit shown in figure, potential of points A and B are respectively
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Q. In the circuit shown in figure, potential of points A and B are respectively
A
$-8V,2V$
B
$\frac{10}{3}V,\frac {20}{3}V $
C
$-6V ,4V$
D
$- \frac{10}{3}V,\frac {20}{3}V $
Solution:
Effective capacitance = $\frac {2C}{5}$
Charge on system = $\frac {20C}{5} = 4C$ According to $K.V.L.$ $V_A + \frac {3q}{2C} = 0$
$V_A = - \frac {3\times q}{2C}\times 4C = -6V$
$and \, V_B=\frac {q}{C}=0\Rightarrow \,V_B=4$
