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Question

Find the potential difference between points $$A$$ and $$B$$ of the system shown in Fig. if the emf is equal to $$\varepsilon = 110\ V$$ and the capacitance ratio $$C_{2}/ C_{1} = \eta = 2.0$$.
1851444_32754b76f4764bf59c7eeb5a386128d0.jpg


Solution

Let us distribute the charges, as shown in the figure.
Now, we know that in a closed circuit, $$-\triangle \varphi = 0$$
So, in the loop, $$DCFED$$,
$$\dfrac {q_{1}}{C_{1}} - \dfrac {q_{2}}{C_{1}} - \dfrac {q_{2}}{C_{2}} = \xi$$ or, $$q_{1} = C_{1} \left [\xi + q_{2} \left (\dfrac {1}{C_{1}} + \dfrac {1}{C_{2}}\right )\right ] .... (1)$$
Again in the loop $$DGHED$$,
$$\dfrac {q_{1}}{C_{1}} + \dfrac {q_{1} + q_{2}}{C_{2}} = \xi .... (2)$$
Using Eqs. (1) and (2), we get
$$q_{2} \left [\dfrac {1}{C_{1}} + \dfrac {3}{C_{2}} + \dfrac {C_{1}}{C_{2}}\right ] = -\dfrac {\xi C_{1}}{C_{2}}$$
Now, $$\varphi_{A} - \varphi_{B} = \dfrac {-q_{2}}{C_{2}} = \dfrac {\xi}{C_{2}^{2}/ C_{1}} \dfrac {1}{\left [\dfrac {1}{C_{1}} + \dfrac {3}{C_{2}} + \dfrac {C_{1}}{C_{2}^{2}}\right ]}$$
or, $$\varphi_{A} - \varphi_{B} = \dfrac {\xi}{\left [\dfrac {C_{2}^{2}}{C_{1}^{2}} + \dfrac {3C_{2}}{C_{1}} + 1\right ]} = \dfrac {\xi}{\eta^{2} + 3\eta + 1} = 10\ V$$.

1784496_1851444_ans_9b100df17f934c3081cdc9788b78f920.png

Physics
NCERT
Standard XII

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