MP Board · Class 12 · Physics · Current ElectricityA potentiometer wire of length $10\text{ m}$ has a resistance of $20\, \Omega$. It is connected in series with a $15\text{ V}$ battery and an external resistance of $40\, \Omega$. Calculate: The potential gradient along the wire. The balancing length for a cell of EMF $1.2\text{ V}$ connected in the secondary circuit.

Given Data:

• Length of the potentiometer wire, $L = 10\text{ m}$
• Resistance of the potentiometer wire, $R_w = 20, \Omega$
• EMF of the battery in primary circuit, $E = 15\text{ V}$
• External resistance in series, $R = 40, \Omega$
• EMF of the cell in secondary circuit, $E' = 1.2\text{ V}$

Step 1: Calculate the total resistance of the primary circuit ($R_{\text{total}}$)\nThe total resistance of the primary circuit is the sum of the external resistance and the resistance of the potentiometer wire:

$$R_{\text{total}} = R + R_w$$ $$R_{\text{total}} = 40, \Omega + 20, \Omega = 60, \Omega$$

Step 2: Calculate the current in the primary circuit ($I$)\nUsing Ohm's law, the current flowing through the primary circuit is:

$$I = \frac{E}{R_{\text{total}}}$$ $$I = \frac{15\text{ V}}{60, \Omega} = 0.25\text{ A}$$

Step 3: Calculate the potential drop across the potentiometer wire ($V_w$)\nThe potential drop across the potentiometer wire is given by:

$$V_w = I \times R_w$$ $$V_w = 0.25\text{ A} \times 20, \Omega = 5\text{ V}$$

Step 4: Calculate the potential gradient along the wire ($k$)\nPotential gradient ($k$) is defined as the potential drop per unit length of the wire:

$$k = \frac{V_w}{L}$$ $$k = \frac{5\text{ V}}{10\text{ m}} = 0.5\text{ V/m}$$

Step 5: Calculate the balancing length for the secondary cell ($l$)\nThe EMF of the secondary cell is balanced by the potential drop across length $l$ of the wire:

$$E' = k \times l$$ $$l = \frac{E'}{k}$$ $$l = \frac{1.2\text{ V}}{0.5\text{ V/m}} = 2.4\text{ m}$$

Final Answer:

1. The potential gradient along the wire is $0.5\text{ V/m}$.
2. The balancing length for the cell is $2.4\text{ m}$.