In order to measure the internal resistance $r_{1}$ of a cell of emf E, a meter bridge of wire resistance $R_{0} = 50 Ω$ , a resistance $R_{0} / 2$ , another cell of emf $E / 2$ (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 $r_{1} =$ _______ $Ω$ . [2021]

Question

Smart Achievers · Accepted Answer

(3)

$Here null point is at 72 cm$

$i (\frac{R_{0}}{2} + 0.28 R_{0}) = \frac{E_{0}}{2}$

$\Rightarrow i \times 0.78 R_{0} = \frac{E_{0}}{2}$

$i = \frac{E_{0}}{2 \times 0.78 R_{0}} = \frac{E_{0}}{r_{1} + \frac{3}{2} R_{0}}$

$or, r_{1} + 1.5 R_{0} = 1.56 R_{0}$

$\Rightarrow r_{1} = 1.56 R_{0} - 1.5 R_{0} = 0.06 R_{0}$

$∴$ $r_{1} = 0.06 \times 50 = 3 Ω$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) Here null point is at 72 cm i(R02+0.28R0)=E02 ⇒i×0.78R0=E02 i=E02×0.78R0=E0r1+32R0 or, r1+1.5R0=1.56R0 ⇒r1=1.56R0-1.5R0=0.06R0 ∴ r1=0.06×50=3 Ω

Want Us to Email you About Special Offers & Updates?