A circuit has a section Ab as shown in the figure with E - 10V- C 1 - 1-0- F- C 2 - 2-0- F and the potential difference V A - V B - 5V- Then determine the voltage across C 1 is -

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Resistors in Series and Parallel

A circuit has...

Question

A circuit has a section Ab as shown in the figure with E = 10V, $C_{1} = 1.0 μ F, C_{2} = 2.0 μ F$ and the potential difference $V_{A} - V_{B} = 5 V$ . Then determine the voltage across $C_{1}$ is :

Zero

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10V

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15V

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Solution

The correct option is B 10V
R.E.F image
Given Date
$E = 10 V$
$C_{1} = 1 μ F$
$C_{2} = 2 μ F$
$V_{A} - V_{B} = 5 V$
$V_{A} - V_{B} = V_{C 1} - \sqrt{2} + V_{C 2}$ $Q = C V = V = \frac{Q}{2}$
$= \frac{q}{C_{1}} - E + \frac{q}{C_{2}}$
$V_{A} - V_{B} + E = q [\frac{1}{C_{1}} + \frac{1}{C_{2}}]$
$= q [\frac{C_{1} + C_{2}}{C_{1} - C_{2}}] ∴ q = \frac{(V_{A} - V_{B} + F_{2}) C_{1} C_{2}}{C_{1} + C_{2}}$
Now Voltage across $C_{1}$
$V_{1} = \frac{q}{C_{1}} = \frac{[V_{A} - V_{B} + E] C_{2}}{C_{1} + C_{2}} = \frac{(5 + 10)^{2}}{1 + 2}$
$= 10 V$
$V_{2} = \frac{q}{C_{2}} = \frac{(V_{A} - V_{B} + E) C_{1}}{C_{1} + C_{2}} = \frac{(5 + 10) 1}{1 + 2} = 5 V$
Now, $V_{1} = 10 v$ $V_{2} = 5 V$
$∴ V o l t a g e a c r o s s V_{1} = 10 V$
$A n s w e r i s C$

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