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V-I Characteristics for a P-N Junction Diode

The resistanc...

Question

The resistance values of the bridge circuit shown in figure are $R_{1} = R_{2} = R_{3} = R and R_{4} = R + Δ R .$ The bridge is balanced by introducing a small voltage v. The value of $Δ R$ is

A

\frac{R V}{E}

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B

\frac{R V}{E - V}

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C

\frac{2 R V}{E - 2 V}

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D

\frac{R V}{E + 2 V}

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Solution

The correct option is C $\frac{2 R V}{E - 2 V}$
Since, the bridge becomes balanced, by introducing 'v' in the arm $R_{3},$ so the current flowing through the galvanometer becomes zero, and it can be shown as a open circuit.

$I_{1} = \frac{E}{R_{1} + R_{2}} = \frac{E}{R + R} = \frac{E}{2 R}$

$I_{2} = \frac{E - V}{R_{3} + R_{4}} = \frac{E - V}{R + Δ R + R}$

$= \frac{E - V}{2 R + Δ R}$

For bridge to be balanced,

$E_{a b} = E_{a d}$

$∴ I_{1} R_{1} = I_{2} R_{4}$

$\frac{E}{2 R} (R) = \frac{E - v}{2 R + Δ R} (R + Δ R)$

$\frac{E}{2} = \frac{(E - V) (R + Δ R)}{(2 R + Δ R)}$

$E (2 R + Δ R) = 2 (E - V) (R + Δ R)$

$2 E R + E (Δ R) = 2 [E R + (Δ R) E - V R - V Δ R]$

$∴ E (Δ R) = 2 E (Δ R) - 2 V R - 2 V (Δ R)$

$2 V R + 2 V (Δ R) = E (Δ R)$

$2 V R = E (Δ R) - 2 V (Δ R)$

$(Δ R) [E - 2 V] = 2 V R$

$Δ R = \frac{2 R V}{E - 2 V}$

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