JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Current Electricity

Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length

l_{1}

is 40 cm. Now an unknown resistance x is connected in series with P and new balancing length is found to be 80 cm measured from the same end. Then the value of x will be ______

Ω

Answer: 20

Solution

In a meter bridge, the balance condition gives

\frac{P}{Q} = \frac{l}{100 - l}

.

Initially with

l_{1} = 40

cm:

\frac{P}{Q} = \frac{40}{60} = \frac{2}{3}

, so

P = 4Ω

and

Q = 6Ω

.

After adding resistance x in series with P, the new balance at 80 cm gives:

\frac{P + x}{Q} = \frac{80}{20} = 4

.

Dividing the second equation by the first:

\frac{P + x}{P} = 4 \times \frac{3}{2} = 6

.

This gives

1 + \frac{x}{P} = 6

, so

\frac{x}{P} = 5

.

Therefore

x = 5 P = 5 \times 4 = 20Ω

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Current Electricity

Solution

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