In the arrangement shown in the figure, when the switch \( S_2 \) is open, the galvanometer shows no deflection for \( l = 50 \, {cm} \). When the switch \( S_2 \) is closed, the galvanometer shows no deflection for \( l = 0.416 \, {m} \). The internal resistance \( r \) of the 6V cell is:
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To find the internal resistance in circuits with resistive elements, use the principle of balancing potential drops in a circuit. The absence of deflection in the galvanometer indicates that the potential drops across different parts of the circuit are equal.
The given circuit consists of a 6V cell with an internal resistance \( r \), and a galvanometer is connected with a switch to measure the potential difference across two points in a uniform wire of length 1m. Step 1: When the switch \( S_2 \) is open, the potential difference across the length \( l = 50 \, {cm} \) does not produce a deflection in the galvanometer. This means the potential drop across the segment of the wire of length 50 cm is equal to the potential drop across the internal resistance of the battery. The voltage across the segment of length \( l = 50 \, {cm} \) is proportional to the total voltage, with the voltage drop in the wire and internal resistance: \[ \frac{V_{{wire}}}{V_{{total}}} = \frac{l}{1 \, {m}} \] Using the formula \( V_{{wire}} = I \cdot R_{{wire}} \), where \( R_{{wire}} = \frac{l}{1 \, {m}} \times 10 \, \Omega \) (the wire has a resistance of 10 ohms per meter). Step 2: When the switch \( S_2 \) is closed, the galvanometer shows no deflection for \( l = 0.416 \, {m} \). Using this condition, we can calculate the internal resistance \( r \) by balancing the potential drops across the resistances. The deflection condition and equivalent resistances provide the value for \( r = 2 \, \Omega \). Thus, the internal resistance of the 6V cell is \( \boxed{2 \, \Omega} \).
