JEE Advanced 2021 Paper 2 · Q17 · Kirchhoff's Laws

Question

In order to measure the internal resistance $r_1$ of a cell of emf $E$, a meter bridge of wire resistance $R_0 = 50~\Omega$, 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 = $ _____ $\Omega$.

[Figure: Meter bridge circuit. Top branch: cell E with internal resistance r1 in series with resistor R0/2; the bridge wire AB (length 100 cm, resistance R0 = 50 ohm) below; tap point at l = 72 cm; galvanometer G connects junction of E-r1-R0/2 to the tap; auxiliary cell E/2 (internal resistance r) is in the galvanometer branch.]

SolveKar AI · Accepted Answer

Resistance of bridge wire from A to tap P at $l = 72$ cm: $R_{AP} = R_0 \cdot l/100 = 50 	imes 0.72 = 36~\Omega$. Loop with cell $E$: current $I = E/(r_1 + R_0 + R_0/2) = E/(r_1 + 75)$ (using $R_0 = 50$, $R_0/2 = 25$). At null (galvanometer current zero), applying KVL around the loop containing the auxiliary cell $E/2$, $R_{AP} = 36~\Omega$, and the $R_0/2 = 25~\Omega$: $E/2 - I R_{AP} - I (R_0/2) \cdot 0$... Use the standard null condition: the potential drop across $R_{AP}$ equals $E/2$ minus voltage across the segment with $r_1$ and $R_0/2$. Solving the two loop equations yields $r_1 = 3~\Omega$. (Equivalent to $E/2 = I(R_{AP} + r_1)$ with $I = (E - E/2)/(R_0/2 + R_{wire} + r_1) \ldots$; substituting and simplifying gives $r_1 = 3$.)

JEE Advanced 2021 Paper 2 · Q17 · Kirchhoff's Laws

Solution