MCQ
The potential difference between the points $P$ and $Q$ in the adjoining circuit will be :-
  • A
    $\frac{{\left( {{C_1}{C_4} - {C_2}{C_3}} \right)E}}{{\left( {{C_1} + {C_3}} \right)\left( {{C_2} + {C_4}} \right)}}$
  • B
    $\frac{{{C_2}{C_3}E}}{{{C_1}{C_2}\left( {{C_3} + {C_4}} \right)}}$
  • $\frac{{\left( {{C_2}{C_3} - {C_1}{C_4}} \right)E}}{{\left( {{C_1} + {C_2}} \right)\left( {{C_3} + {C_4}} \right)}}$
  • D
    $\frac{{\left( {{C_2}{C_3} - {C_1}{C_4}} \right)E}}{{\left( {{C_1} + {C_2} + {C_3} + {C_4}} \right)}}$

Answer

Correct option: C.
$\frac{{\left( {{C_2}{C_3} - {C_1}{C_4}} \right)E}}{{\left( {{C_1} + {C_2}} \right)\left( {{C_3} + {C_4}} \right)}}$
c
$\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series, charge on each will remain same.

$\left(\mathrm{V}_{\mathrm{P}}-\mathrm{O}\right) \cdot \mathrm{C}_{1}=\left(\mathrm{E}-\mathrm{V}_{\mathrm{P}}\right) \mathrm{C}_{2}$

$\mathrm{V}_{\mathrm{P}}=\frac{\mathrm{C}_{2} \mathrm{E}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

$\mathrm{C}_{3} $ and $ \mathrm{C}_{4}$ are in series,

charge on each will remain same,

$\left(\mathrm{V}_{\mathrm{Q}}-\mathrm{0}\right) \cdot \mathrm{C}_{3}=\left(\mathrm{E}-\mathrm{V}_{\mathrm{Q}}\right) \mathrm{C}_{4}$

$\mathrm{V}_{\mathrm{Q}}=\frac{\mathrm{C}_{4} \mathrm{E}}{\mathrm{C}_{3}+\mathrm{C}_{4}}$

Hence $\mathrm{V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}=\frac{\left(\mathrm{C}_{2} \mathrm{C}_{3}-\mathrm{C}_{1} \mathrm{C}_{4}\right) \mathrm{E}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)\left(\mathrm{C}_{3}+\mathrm{C}_{4}\right)}$

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