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 $$C_{2}/ C_{1} = \eta = 2.0$$.

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, $$\varphi_{A} - \varphi_{B} = \dfrac {\xi}{\left [\dfrac {C_{2}^{2}}{C_{1}^{2}} + \dfrac {3C_{2}}{C_{1}} + 1\right ]} = \dfrac {\xi}{\eta^{2} + 3\eta + 1} = 10\ V$$.PhysicsNCERTStandard XII

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More