A meter bridge with two resistances R1 and R2 as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when a 16 resistance is…

Given: A meter bridge is balanced at 40 cm from point P initially. After connecting 16 in parallel with R2, the null point shifts to 50 cm from P. Find: The values of R1 and R2. In a meter bridge at balance condition, R1R2 = l100-l Step 1: Initial balance condition For null point at 40 cm, R1R2 = 4060 = 23 So, R1 = 23R2 Step 2: New balance condition after adding 16 in parallel with R2 The equivalent resistance is R2' = 16R216 + R2 For the new null point at 50 cm, R1R2' = 5050 = 1 Hence, R1 = R2' Step 3: Substitute R1 = 23R2 23R2 = 16R216 + R2 23(16 + R2) = 16 32 + 2R2 = 48 R2 = 8 Now, R1 = 23 8 = 163 Therefore, the correct option is C and the resistances are R2 = 8 and R1 = 163.

A meter bridge with two resistances R1 and R2 as shown in figure… — Solution | JEE Physics

MCQMediumJEE 2026Wheatstone Bridge & Meter Bridge

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at $40 \, \text{cm}$ from the point $P$ . The null point changed to $50 \, \text{cm}$ from the point $P$ , when a $16 \, \Omega$ resistance is connected in parallel to $R_2$ . The values of resistances $R_1$ and $R_2$ are

Meter bridge circuit with resistances R1 and R2 in the two gaps, wire PQ of length 100 cm, and null point measured from point P.

A
$R_2 = 4 \, \Omega,\; R_1 = \frac{4}{3} \, \Omega$
B
$R_2 = 16 \, \Omega,\; R_1 = \frac{16}{3} \, \Omega$
C
$R_2 = 8 \, \Omega,\; R_1 = \frac{16}{3} \, \Omega$
D
$R_2 = 12 \, \Omega,\; R_1 = \frac{12}{3} \, \Omega$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A meter bridge is balanced at $40 \, \text{cm}$ from point $P$ initially. After connecting $16 \, \Omega$ in parallel with $R_2$ , the null point shifts to $50 \, \text{cm}$ from $P$ .

Find: The values of $R_1$ and $R_2$ .

In a meter bridge at balance condition,

\frac{R_1}{R_2} = \frac{l}{100-l}

Step 1: Initial balance condition

For null point at $40 \, \text{cm}$ ,

\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3}

So,

R_1 = \frac{2}{3}R_2

Step 2: New balance condition after adding $16 \, \Omega$ in parallel with $R_2$

The equivalent resistance is

R_2' = \frac{16R_2}{16 + R_2}

For the new null point at $50 \, \text{cm}$ ,

\frac{R_1}{R_2'} = \frac{50}{50} = 1

Hence,

R_1 = R_2'

Step 3: Substitute $R_1 = \frac{2}{3}R_2$

\frac{2}{3}R_2 = \frac{16R_2}{16 + R_2}

\frac{2}{3}(16 + R_2) = 16

32 + 2R_2 = 48

R_2 = 8 \, \Omega

Now,

R_1 = \frac{2}{3} \times 8 = \frac{16}{3} \, \Omega

Therefore, the correct option is C and the resistances are $R_2 = 8 \, \Omega$ and $R_1 = \frac{16}{3} \, \Omega$ .

Common mistakes

Using the meter bridge balance condition incorrectly as $\frac{R_2}{R_1} = \frac{l}{100-l}$ . This reverses the resistance ratio and gives wrong values. Use the same arm convention consistently: here $\frac{R_1}{R_2} = \frac{l}{100-l}$ .
Treating the added $16 \, \Omega$ resistor as series combination with $R_2$ . It is connected in parallel, so the equivalent resistance must decrease. Use $R_2' = \frac{16R_2}{16+R_2}$ instead.
Using the new balance length $50 \, \text{cm}$ incorrectly. At $50 \, \text{cm}$ , the ratio is $\frac{50}{50} = 1$ , so the two effective bridge arm resistances are equal. Do not substitute $50$ directly as a resistance value.

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