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प्रश्न
Find the potential difference between the points A and B and between the points B and C of the figure in steady state.
उत्तर
Equivalent capacitance of the circuit can be calculatesd as,
Ceq:-
\[1 \mu F\text{ is in parallel with, }C = \frac{1}{\frac{1}{(3 + 3)} + \frac{1}{(1 + 1)}} = \frac{3}{2}\mu F\]
So, \[C_{eq} = \frac{3}{2} + 1 = \frac{5}{2} \mu F\]
V = 100 V
Total charge in the circuits is,
\[Q = C_{eq} V = \frac{5}{2} \times 100 = 250 \mu c\]
As the volatge across 1μf is 100 V, therfore charge stored on 1 μf capacitors = 100 μC
Charge flowing from A to B = (250 - 100) = 150 μC
Ceq between A and B is 6 μf.
Potential drop across AB,
\[V = \frac{Q}{C}\]
\[ \Rightarrow V_{AB} = \frac{150}{6} = 25 V\]
Potential drop across BC = (100 - 25) = 75 V
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