First slide

meter bridge

Question

In a metre bridge [Fig], the gaps are closed by two resistances P and Q and the balance point is obtained
at 40 cm. When Q is shunted by resistance of 10 $Ω$ , the balance point shifts to 50 cm. The values of P and Q are

NA

A

$(10 / 3) Ω, 5 Ω$
B

$20 Ω, 30 Ω$
C

$10 Ω, 15 Ω$
D

$5 Ω, (15 / 2) Ω$

Solution

For the balance of metre bridge
$\frac{P}{Q} = \frac{R}{S} = \frac{40}{60} = \frac{2}{3}$
or $P = \frac{2}{3} Q$ ……….(1)
When Q is shunted by a resistance of 10 $Ω$ then net resistance in this gap should be a parallel combination of Q and 10 $Ω$ resistance i.e., 10Q /10 +Q).
Now the balance shifts to 50 cm i.e., R = 50cm and S = 50 cm. In this case
$\frac{P}{[10 Q / (10 + Q)]} = \frac{50}{50} = 1$ ……..(2)
Substituting the value of P from eq. (1) in eq. (2), we get
$\frac{(2 / 3) Q}{[10 Q / (10 + Q)]} = 1$
Solving we get $Q = 5 Ω$
From eq. (1), $P = \frac{2}{3} \times 5 = \frac{10}{3} Ω$

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