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Meter Bridge

Consider a Wh...

Question

Consider a Wheatstone bridge with resistance and capacitance connected as shown.
Find the condition on the resistance and the capacitance such that the bridge remains balanced at all times.

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Solution

Suppose that the bridge is balanced ie., $V_{A B} = V_{A D}; V_{B C} = V_{D C}$
Let the current and the charges on the capacitors in the circuit as shown then
$i_{1} R_{1} = i_{3} R_{3} . . . (i)$
$\frac{q_{2}}{C_{2}} = \frac{q_{4}}{C_{4}} . . . (i i)$
Consider the charging of the part of the circuit shown alongside. Let $q_{2}$ be the charge on $C_{2}$ and $i_{1}$ be the current in the circuit
Then,
$i_{1} R_{1} + \frac{q_{2}}{C_{2}} = ε . . . (i i i); \frac{d q_{2}}{d t} + \frac{q_{2}}{R_{2} C_{2}} = i_{1} . . . (i v)$
Substituting (iv) in (ii) and simplifying
$\frac{d q_{2}}{d t} + \frac{q_{2}}{R_{e q} C_{2}} = \frac{ε}{R_{1}} [\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}]$
$∴ q_{2} = \frac{ε R_{e q} C_{2}}{R_{1}} ⎡ ⎢ ⎢ ⎣ 1 - e^{\frac{1}{R_{e q} C_{2}}} ⎤ ⎥ ⎥ ⎦ = \frac{ε R_{2} C_{2}}{R_{1} + R_{2}} [1 - e^{- t / R_{e q} C_{2}}]$
Similarly for the other circuit, we have
$q_{4} = \frac{ε R_{4} C_{4}}{R_{3} + R_{4}} ⎡ ⎢ ⎢ ⎣ 1 - e^{\frac{1}{R_{e q}^{'} C_{2}}} ⎤ ⎥ ⎥ ⎦ [\frac{1}{R_{e q}^{'}} = \frac{1}{R_{3}} + \frac{1}{R_{4}}]$
Now $\frac{q_{2}}{C_{2}} = \frac{q_{4}}{C_{4}}$ which leads to
$\frac{R_{2}}{R_{1} + R_{2}} = \frac{R_{4}}{R_{3} + R_{4}}; \frac{1}{R_{e q} C_{2}} = \frac{1}{R_{e q}^{'} C_{4}} o r \frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}; R_{1} C_{2} = R_{3} C_{4}$

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