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

Balanced condition for 1st case:R1R2=l100−l=50100−50=1 so R1=R2In 2nd case : when 24 Ω is connected in parallel to R2, the resistance of R2 arm becomes RP=24R224+R2Now balance condition is R1RP=70100−70⇒R1RP=7/3⇒R2=(7/3)RP as R1=R2⇒ R2=(7/3)24R224−R2⇒ 24R2+R22=56R2⇒R22=32R2⇒ R2=32Ω

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

Balanced condition for 1st case:R1R2=l100−l=50100−50=1 so R1=R2In 2nd case : when 24 Ω is connected in parallel to R2, the resistance of R2 arm becomes RP=24R224+R2Now balance condition is R1RP=70100−70⇒R1RP=7/3⇒R2=(7/3)RP as R1=R2⇒ R2=(7/3)24R224−R2⇒ 24R2+R22=56R2⇒R22=32R2⇒ R2=32Ω

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

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$2 Ω$

$8 Ω$

$12 Ω$

$32 Ω$

answer is D.

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Detailed Solution

Balanced condition for 1st case: $\frac{R_{1}}{R_{2}} = \frac{l}{100 - l} = \frac{50}{100 - 50} = 1$

$so R_{1} = R_{2}$

In 2nd case : when 24 Ω is connected in parallel to R₂, the resistance of R₂ arm becomes

$R_{P} = \frac{24 R_{2}}{24 + R_{2}}$

Now balance condition is $\frac{R_{1}}{R_{P}} = \frac{70}{100 - 70}$

$\begin{array}{l} ⇒ \frac{R_{1}}{R_{P}} = 7 / 3 \\ ⇒ R_{2} = (7 / 3) R_{P} as R_{1} = R_{2} \\ ⇒ R_{2} = (7 / 3) \frac{24 R_{2}}{24 - R_{2}} \\ ⇒ 24 R_{2} + R_{2}^{2} = 56 R_{2} \\ ⇒ R_{2}^{2} = 32 R_{2} \\ ⇒ R_{2} = 32 Ω \end{array}$