First slide

Electrical measuring instruments

Question

Two resistances are connected in two gaps of a meter bridge. The balance point is 20cm from the zero end. A resistance of $15 Ω$ is connected in series with the smaller of the two. The null point shifts to 40cm. Then value of the smaller resistance is:

Moderate

A

$3 Ω$
B

$6 Ω$
C

$9 Ω$
D

$12 Ω$

Solution

Given
Balancing length $l = 20 cm$
Resistance= $15 Ω$
If the unknown resistances are P and Q, then for first case we have at null point 20cm
$\frac{P}{Q} = \frac{20}{100 - 20} or Q = 4 P$
$\Rightarrow P < Q$
In second case we have null point at 40cm
$\begin{array}{l} \frac{P + 15}{Q} = \frac{40}{100 - 40} \\ \Rightarrow \frac{P + 15}{Q} = \frac{2}{3} \end{array}$
Solving above equations we get, $P = 9 Ω$

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