Question

In a meter bridge experiment when a resistance wire is connected in the left gap, the balance point is found at the 30^(th) cm. When the wire is replaced by another wire,t he balance point is found at the 60^(th) cm. Find the balance point when the two wires connected in series and in parallel in the left gap successively

Answer 1

Answer:

When two wires are in series

L = 65.8 cm

When two wires are in parallel

L = 25 cm

Explanation:

As we know that the meter bridge is based upon the principle of Wheatstone Bridge

So here we can say

$R = \frac{L}{100 - L} S$

so here for first resistance we have

$R_1 = \frac{30}{100 - 30}(S) = \frac{3}{7}S$

for second resistance we have

$R_2 = \frac{60}{100 - 60}(S) = \frac{3}{2}S$

now when two resistors are in series

$R_{net} = R_1 + R_2$

$R_{net} = \frac{3}{7}S + \frac{3}{2}S$

$R_{net} = \frac{27}{14}S$

so we have

$\frac{L}{100 - L}S = \frac{27}{14} S$

$14L = 2700 - 27 L$

$L = 65.8 cm$

Now when resistors are connected in parallel

$\frac{1}{R_{net}} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_{net}} = \frac{7}{3S} + \frac{2}{3S}$

$R_{net} = \frac{S}{3}$

now it is

$\frac{L}{100 - L} S = \frac{S}{3}$

$3L = 100 - L$

$L = 25 cm$

#Learn

Topic : Meter Bridge

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