Current Electricity - Potentiometer and Wheatstone Bridge

In a Metre Bridge, the balance point is found at a distance of $40\text{ cm}$ from end $A$ when a resistor $R = 12 \Omega$ is in the left gap and an unknown resistor $S$ is in the right gap. Calculate the value of $S$.

Given $l = 40\text{ cm}$ and $R = 12 \Omega$. Using the Metre Bridge formula: $S = \frac{(100 - l)}{l} R$. Substituting the values: $S = \frac{(100 - 40)}{40} \times 12 = \frac{60}{40} \times 12 = 1.5 \times 12 = 18 \Omega$.

The Metre Bridge works on the Wheatstone principle where the ratio of resistances equals the ratio of the lengths of the wire segments.

A potentiometer wire has a length of $4\text{ m}$ and a resistance of $8 \Omega$. A cell of EMF $2\text{ V}$ is connected across it. Calculate the potential gradient.

Length $L = 4\text{ m}$, Resistance $R_w = 8 \Omega$, $V = 2\text{ V}$. Potential gradient $k = \frac{V}{L} = \frac{2\text{ V}}{4\text{ m}} = 0.5\text{ V/m}$. To convert to $\text{V/cm}$: $k = \frac{0.5}{100} = 0.005\text{ V/cm}$.

Potential gradient is the potential drop divided by the total length of the potentiometer wire.

With a cell of EMF $E$ in the secondary circuit of a potentiometer, the balance point is $200\text{ cm}$. When a resistor of $5 \Omega$ is connected across the cell, the balance point shifts to $150\text{ cm}$. Find the internal resistance of the cell.

Given $l_1 = 200\text{ cm}$, $l_2 = 150\text{ cm}$, and external resistance $R = 5 \Omega$. Internal resistance $r = R \left( \frac{l_1}{l_2} - 1 \right)$. $r = 5 \left( \frac{200}{150} - 1 \right) = 5 \left( \frac{4}{3} - 1 \right) = 5 \times \frac{1}{3} \approx 1.67 \Omega$.

The internal resistance is calculated by comparing the balancing length of the cell in open circuit ($l_1$) and closed circuit ($l_2$).

In a Metre Bridge experiment, the balance point is obtained at $l = 60\text{ cm}$ from the left end. If the resistance in the left gap is $15 \Omega$, determine the value of the unknown resistance $X$ in the right gap. If the resistors are interchanged, what will be the new balance length?

1. Using the Wheatstone Bridge principle for a Metre Bridge: $\frac{R}{X} = \frac{l}{100 - l}$ $15/X = 60 / (100 - 60)$ $15/X = 60 / 40 = 3/2$ $X = (15 \times 2) / 3 = 10 \Omega$

2. When resistors are interchanged, the new balance length $l'$ is: $\frac{X}{R} = \frac{l'}{100 - l'}$ $10/15 = l' / (100 - l')$ $2/3 = l' / (100 - l')$ $200 - 2l' = 3l' \Rightarrow 5l' = 200$ $l' = 40\text{ cm}$

The ratio of resistances in the gaps equals the ratio of the lengths of the wire segments. Interchanging the resistors simply swaps the segments, moving the balance point from $l$ to $100 - l$.

A potentiometer has a wire of length $10\text{ m}$ and resistance $20 \Omega$. It is connected in series with a $2\text{ V}$ battery and a $30 \Omega$ external resistor. Find the potential gradient of the wire and the EMF of a cell that balances at $600\text{ cm}$.

1. Total resistance of primary circuit: $R_{total} = R_{wire} + R_{ext} = 20 + 30 = 50 \Omega$

2. Current in the circuit: $I = V / R_{total} = 2 / 50 = 0.04\text{ A}$

3. Potential drop across wire ($V_w$): $V_w = I \times R_{wire} = 0.04 \times 20 = 0.8\text{ V}$

4. Potential gradient ($k$): $k = V_w / L = 0.8 / 10 = 0.08\text{ V/m}$

5. EMF of the cell ($E$): $E = k \times l = 0.08 \times 6 = 0.48\text{ V}$ Note: $600\text{ cm} = 6\text{ m}$.

First find the total resistance of the primary circuit to calculate the current. Use this current to find the potential drop across the potentiometer wire, then divide by its length to get the gradient.