In a meter bridge experiment to determine the value of unknown resistance, first the resistances 2 and 3 are connected in the left and right gaps of the bridge and the null point is obtained at a…

Given: Left gap resistance is 2 and right gap resistance is 3 initially. The null point is at distance l from the left end. When an unknown resistance x is connected in parallel with 3, the null point shifts 10 cm to the right. Find: The value of x. For a meter bridge at balance, R leftR right = l100-l Initially, 23 = l100-l So, 2(100-l)=3l 200-2l=3l 5l=200 l=40 cm The null point shifts 10 cm to the right, therefore l' = 40 + 10 = 50 cm Let the new equivalent resistance in the right gap be R. Since x is connected in parallel with 3, R = 3x3+x Applying the balance condition again, 2R = 50100-50 2R = 5050 = 1 Hence, R=2 Now solve 3x3+x=2 3x=6+2x x=6 Therefore, the value of the unknown resistance is 6.

In a meter bridge experiment to determine the value of unknown re… — Solution | JEE Physics

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In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2\,\Omega$ and $3\,\Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l$ cm from the left end. Now, when an unknown resistance $x\,\Omega$ is connected in parallel to $3\,\Omega$ , the null point is shifted by $10\,cm$ to the right. The value of $x$ is _____ $\Omega$ .

Answer

Correct answer:6

Step-by-step solution

Standard Method

Given: Left gap resistance is $2\,\Omega$ and right gap resistance is $3\,\Omega$ initially. The null point is at distance $l$ from the left end. When an unknown resistance $x\,\Omega$ is connected in parallel with $3\,\Omega$ , the null point shifts $10\,cm$ to the right.

Find: The value of $x$ .

For a meter bridge at balance,

\frac{R_\text{left}}{R_\text{right}} = \frac{l}{100-l}

Initially,

\frac{2}{3} = \frac{l}{100-l}

So,

2(100-l)=3l

200-2l=3l

5l=200

l=40\,\text{cm}

The null point shifts $10\,cm$ to the right, therefore

l' = 40 + 10 = 50\,\text{cm}

Let the new equivalent resistance in the right gap be $R$ . Since $x$ is connected in parallel with $3\,\Omega$ ,

R = \frac{3x}{3+x}

Applying the balance condition again,

\frac{2}{R} = \frac{50}{100-50}

\frac{2}{R} = \frac{50}{50} = 1

Hence,

R=2

Now solve

\frac{3x}{3+x}=2

3x=6+2x

x=6

Therefore, the value of the unknown resistance is $6\,\Omega$ .

Common mistakes

Using the wrong meter bridge ratio. The balance condition must be written as resistance ratio equals wire length ratio, that is $\frac{R_1}{R_2}=\frac{l}{100-l}$ . Reversing this ratio gives an incorrect value of $l$ and spoils the rest of the solution.
Treating the shift of $10\,cm$ to the right as the new null length directly. The original null point is first found to be $40\,cm$ , so the new position becomes $50\,cm$ . Always update the position relative to the initial balance point.
Adding the parallel resistances incorrectly. The combination of $3\,\Omega$ and $x\,\Omega$ is not $3+x$ . For parallel connection, the equivalent resistance is $\frac{3x}{3+x}$ , which must be less than each individual resistance.

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