Current Electricity - Kirchhoff's Laws and Electrical Measurements

In a Meter Bridge, the null 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 Meter Bridge formula: $S = R \left( \frac{100 - l}{l} \right)$. Substituting the values: $S = 12 \times \left( \frac{100 - 40}{40} \right) = 12 \times \frac{60}{40} = 12 \times 1.5 = 18 \, \Omega$.

The unknown resistance $S$ is found by applying the balanced Wheatstone Bridge condition adapted for the lengths of the wire.

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

Total length $L = 4 \text{ m}$, Resistance $R = 8 \, \Omega$, Voltage $V = 2 \text{ V}$. Potential gradient $k = \frac{V}{L}$. So, $k = \frac{2 \text{ V}}{4 \text{ m}} = 0.5 \text{ V/m}$.

Potential gradient is the potential drop per unit length of the potentiometer wire, which determines the sensitivity of the instrument.

In a potentiometer experiment, the balancing length for a cell in open circuit is $350 \text{ cm}$. When a resistor of $10 \, \Omega$ is connected across the cell, the balancing length shifts to $300 \text{ cm}$. Find the internal resistance $r$ of the cell.

Given $l_1 = 350 \text{ cm}$, $l_2 = 300 \text{ cm}$, and $R = 10 \, \Omega$. Using the formula $r = R \left( \frac{l_1}{l_2} - 1 \right)$: $r = 10 \left( \frac{350}{300} - 1 \right) = 10 \left( \frac{7}{6} - 1 \right) = 10 \times \frac{1}{6} \approx 1.67 \, \Omega$.

The internal resistance is calculated by comparing the balancing lengths of the cell in an open circuit (EMF) and a closed circuit (Terminal Voltage).

Determine the current flowing through the $4 \, \Omega$ resistor in the following circuit using Kirchhoff's Laws. The circuit consists of two loops with batteries $E_1 = 10 \text{ V}$ and $E_2 = 12 \text{ V}$ and resistors $R_1 = 2 \, \Omega$, $R_2 = 4 \, \Omega$, and $R_3 = 2 \, \Omega$.

Let $I_1$ be the current from $E_1$ and $I_2$ be the current from $E_2$. At the central branch, the current is $I_1 + I_2$. Applying Loop Rule to Loop 1 (left): $10 - 2I_1 - 4(I_1 + I_2) = 0 \implies 6I_1 + 4I_2 = 10 \dots (1)$ Applying Loop Rule to Loop 2 (right): $12 - 2I_2 - 4(I_1 + I_2) = 0 \implies 4I_1 + 6I_2 = 12 \dots (2)$ Multiplying (1) by 3 and (2) by 2: $18I_1 + 12I_2 = 30$ $8I_1 + 12I_2 = 24$ Subtracting: $10I_1 = 6 \implies I_1 = 0.6 \text{ A}$. Substitute $I_1$ in (1): $6(0.6) + 4I_2 = 10 \implies 3.6 + 4I_2 = 10 \implies 4I_2 = 6.4 \implies I_2 = 1.6 \text{ A}$. Current through $4 \, \Omega$ resistor: $I = I_1 + I_2 = 0.6 + 1.6 = 2.2 \text{ A}$.

We used Kirchhoff's Voltage Law (KVL) for two independent loops and Kirchhoff's Current Law (KCL) at the junction to solve for the individual branch currents.

In the circuit shown, two batteries $E_1 = 6\text{ V}$ and $E_2 = 12\text{ V}$ are connected with three resistors $R_1 = 2\text{ }\Omega$, $R_2 = 4\text{ }\Omega$, and $R_3 = 6\text{ }\Omega$. Using Kirchhoff's laws, calculate the current $I_3$ flowing through the resistor $R_3$.

Let the current from $E_1$ be $I_1$ and from $E_2$ be $I_2$. At the junction, they combine so that $I_3 = I_1 + I_2$.

Applying Kirchhoff's Voltage Law (KVL) to the left loop (clockwise): $-E_1 + I_1 R_1 + (I_1 + I_2) R_3 = 0$ $-6 + 2I_1 + 6(I_1 + I_2) = 0$ $8I_1 + 6I_2 = 6 \quad \text{--- (1)}$

Applying KVL to the right loop (counter-clockwise): $-E_2 + I_2 R_2 + (I_1 + I_2) R_3 = 0$ $-12 + 4I_2 + 6(I_1 + I_2) = 0$ $6I_1 + 10I_2 = 12 \quad \text{--- (2)}$

Multiplying (1) by 3 and (2) by 4: $24I_1 + 18I_2 = 18$ $24I_1 + 40I_2 = 48$

Subtracting the first from the second: $22I_2 = 30 \implies I_2 = \frac{15}{11}\text{ A}$

Substitute $I_2$ into (1): $8I_1 + 6\left(\frac{15}{11}\right) = 6$ $8I_1 = 6 - \frac{90}{11} = \frac{66 - 90}{11} = -\frac{24}{11}$ $I_1 = -\frac{3}{11}\text{ A}$

Calculate $I_3$: $I_3 = I_1 + I_2 = -\frac{3}{11} + \frac{15}{11} = \frac{12}{11} \approx 1.09\text{ A}$

We apply Kirchhoff's First Law at the central node and Kirchhoff's Second Law to the two independent loops. The negative sign for $I_1$ simply indicates that the current actually flows in the opposite direction to our initial assumption.