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Standard XII

Physics

Meter Bridge

The resistanc...

Question

The resistance in the left and right gaps of a balanced meter bridge are $R_{1}$ , $R_{2}$ . The balance point is at 50 cm. If a resistance of $24 Ω$ is connected in parallel to $R_{2}$ the balance point is at 70 cm. The value of $R_{1}$ is

2 Ω

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8 Ω

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32 Ω

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17 Ω

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Solution

The correct option is C $32 Ω$
$\frac{R_{1}}{R_{2}} = \frac{50}{100 - 50} = \frac{50}{50} = 1; R_{1} = R_{2} \frac{R_{1}}{(\frac{24 R_{2}}{24 + R_{2}})} = \frac{70}{100 - 70} = \frac{70}{30} \frac{24 + R_{2}}{24} = \frac{7}{3} 24 + R_{2} = 56 R_{2} = 32 Ω$

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Similar questions

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The resistance in the left and right gaps of a balanced meter bridge are R1, R2. The balance point is at 50 cm. If a resistance of 24Ω is connected in parallel to R2the balance point is at 70 cm. The value of R1 is

The correct option is C 32ΩR1R2=50100−50=5050=1;R1=R2R1(24R224+R2)=70100−70=703024+R224=7324+R2=56R2=32Ω

The resistance in the left and right gaps of a balanced meter bridge are $R_{1}$ , $R_{2}$ . The balance point is at 50 cm. If a resistance of $24 Ω$ is connected in parallel to $R_{2}$ the balance point is at 70 cm. The value of $R_{1}$ is

The correct option is C $32 Ω$
$\frac{R_{1}}{R_{2}} = \frac{50}{100 - 50} = \frac{50}{50} = 1; R_{1} = R_{2} \frac{R_{1}}{(\frac{24 R_{2}}{24 + R_{2}})} = \frac{70}{100 - 70} = \frac{70}{30} \frac{24 + R_{2}}{24} = \frac{7}{3} 24 + R_{2} = 56 R_{2} = 32 Ω$