TS EAMCET · Physics · Current Electricity
When the right gap of a meter bridge consists of two equal resistors in series, the balancing point is at 50 cm. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap, the balancing point is at
- A 60 cm
- B 33.3 cm
- C 25 cm
- D 40 cm
Answer & Solution
Correct Answer
(D) 40 cm
Step-by-step Solution
Detailed explanation
\(\frac{R_L}{2R} = \frac{50}{100 - 50} \Rightarrow R_L = 2R\) \(R_L' = \frac{R_L \cdot R}{R_L + R} = \frac{2R \cdot R}{2R + R} = \frac{2}{3}R\) \(R_R' = R\) \(\frac{R_L'}{R_R'} = \frac{l}{100 - l}\) \(\frac{(2/3)R}{R} = \frac{l}{100 - l}\) \(\frac{2}{3} = \frac{l}{100 - l}\)…
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