In order to measure the internal resistance r 1 of a cell of emf ε , a meter bridge of wire resistance R 0 = 50 Ω , a resistance R 0 2 , another cell of emf ε 2 (internal resistance r ) and a galvanom

Question

In order to measure the internal resistance $r_{1}$ of a cell of emf $ε,$ a meter bridge of wire resistance $R_{0} = 50 Ω$ , a resistance $\frac{R_{0}}{2}$ , another cell of emf $\frac{ε}{2}$ (internal resistance $r$ ) and a galvanometer $G$ are used in a circuit, as shown in the figure. If the null point is found at $l = 72 cm,$ then the value of $r_{1} =____Ω .$

TestHub · Accepted Answer

Resistance of potential wire is R 0 = 50 Ω Resistance of 100 cm wire = 50 Ω So, Resistance of 72 cm wire = 50 100 × 72 = 36 Ω Current, I = ε 2 14 + 25 = ε r 1 + 75 ⇒ r 1 = 3 Ω

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