JEE Advance - Physics (2021 - Paper 2 Online - No. 17)
In order to measure the internal resistance r1 of a cell of emf E, a meter bridge of wire resistance R0 = 50$$\Omega$$, a resistance R0/2, 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 r1 = __________ $$\Omega$$.
Answer
3
Explanation
Resistance of potential wire, R0 = 50$$\Omega$$
Resistance of 100 m wire = 50$$\Omega$$
So, resistance of 72 cm wire $$ = {{50} \over {100}} \times 72 = 36\Omega $$
Current, $$I = {{{\varepsilon \over 2}} \over {14\Omega + 25\Omega }} = {{{\varepsilon \over 2}} \over {{r_1} + 36\Omega }}$$
$$ \Rightarrow {r_1} = 39 - 36 = 3\Omega $$
Comments (0)
