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In the figure below, what is the potential difference between the point A and B and between B and C respectively in steady state

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V_AB=75 V, V_BC=25 V

V_AB=25 V ,V_BC=75 V

V_AB=V_BC=50 V

$V_{A B} = V_{B C} = 100 V$

answer is C.

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Detailed Solution

${C_{eq}} = \frac{{(3 + 3) \times (1 + 1)}}{{(3 + 3) + (1 + 1)}} + 1 = \left( {\frac{{6 \times 2}}{{6 + 2}}} \right)\, + 1$

$= \frac{5}{2}\mu F$

$\therefore Q=C \times V = \frac{5}{2} \times 100 = 250\mu C$

Charge in 6μF branch $= VC = \left( {\frac{{6 \times 2}}{{6 + 2}}} \right)\,100 = 150\mu C$

Question Image

${V_{AB}} = \frac{{150}}{6} = 25\,V\,and\,\,{V_{BC}} = 100 - {V_{AB}} = 75\,V$

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