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Wheatstone Bridge

In the follow...

Question

In the following circuit, each resistor has a resistance of $15 Ω$ and the battery has an e.m.f. of $12 V$ with negligible internal resistance.
When a resistor of resistance $R$ is connected between $D & F$ , no current flows through the galvanometer (not shown in the figure) connected between $C & F$ . Calculate the value of $R$ .

10 Ω

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15 Ω

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5 Ω

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30 Ω

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Solution

The correct option is A $10 Ω$
Voltage at point C $\Rightarrow \frac{V_{c} - 0}{15} = \frac{12 - V_{c}}{30}$
$2 V_{c} = 12 - V_{c}$
$3 V_{c} = 12$
$V_{c} = 4 v o l t$
It is given that there is no any current through line CF $\Rightarrow V_{c} = V_{F}$
$\Rightarrow V_{F} = 4 v o l t$
$\Rightarrow I_{2} = \frac{12 - V_{F}}{15} = \frac{12 - 4}{15} = \frac{8}{15} A$
Now ohm's law for unknown resistance $R_{1} \Rightarrow Δ V = I R_{1}$
$(4 - 0) = (\frac{30}{30 + R_{1}}) (\frac{8}{15}) R_{1}$
$4 = \frac{16 R_{1}}{30 + R_{1}}$
$120 + 4 R_{1} = 16 R_{1}$
$12 R_{1} = 120$
$R_{1} = 10 Ω$

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