Question

Write Kirchhoff’s law for electrical circuits. In the given balanced Wheatstone bridge, find the potential at points B and D and the values of current \(i_1\) and \(i_2\). 

Hint:

In a balanced Wheatstone bridge, the ratio of resistances on both sides is equal, and no current flows through the galvanometer. The current is divided between the two parallel branches of the bridge.

Answer 1

Step 1: Kirchhoff’s Laws.
- Kirchhoff's Current Law (KCL): The sum of currents entering a junction equals the sum of currents leaving the junction.
\[ \sum I_{\text{in}} = \sum I_{\text{out}}. \] - Kirchhoff's Voltage Law (KVL): The sum of the electrical potential differences (voltages) around any closed loop or circuit is zero.
\[ \sum V = 0. \] Step 2: Wheatstone Bridge.
In a Wheatstone bridge, there are four resistors arranged in a diamond shape. The bridge is balanced when the ratio of resistances in one pair is equal to the ratio of resistances in the other pair:
\[ \frac{R_1}{R_2} = \frac{R_3}{R_4}. \] The current in the bridge is divided between two parallel branches. The current \(i\) is divided into \(i_1\) and \(i_2\) at point \(A\), where: \[ i = i_1 + i_2. \] Step 3: Using Kirchhoff's Laws to find the potential and currents.
Using KCL and KVL, we can write the equations for the current through each resistor. For the balanced Wheatstone bridge, the voltage drop across the bridge is zero, so the potential at points B and D is the same, and hence no current flows through the galvanometer \(G\). Therefore, the currents \(i_1\) and \(i_2\) are: \[ i_1 = \frac{V}{R_1 + R_2}, i_2 = \frac{V}{R_3 + R_4}. \] where \(V\) is the applied voltage. Since the bridge is balanced, the potential at points B and D is the same, and no current flows through the galvanometer. Final Answer: The potential at points B and D is the same, and the values of current \(i_1\) and \(i_2\) can be calculated using the above equations based on the resistances in the circuit.