Wheatstone Bridge

In a Wheatstone bridge, there are four resistances , and a galvanometer is connected between the junctions of & and & . The bridge is said to be balanced when the galvanometer shows zero current, i.e., there is no potential difference across it. The relation for balance condition is: In this condition, the ratios of the resistances in the two halves of the bridge are equal.

Interference of Light:

Interference is the phenomenon that occurs when two or more light waves superpose to form a resultant wave. Depending on the phase relationship between the waves, the interference can be constructive or destructive.

Condition for Constructive Interference:

When the path difference between the waves is an integer multiple of the wavelength , the waves interfere constructively:

Condition for Destructive Interference:

When the path difference is an odd multiple of half the wavelength , the waves interfere destructively:

Kirchhoff's law consists of two rules that apply to the current and potential difference in electrical circuits: 1. Kirchhoff's Current Law (KCL): The total current entering a junction equals the total current leaving the junction. Mathematically:
2. Kirchhoff's Voltage Law (KVL): The sum of the electrical potential differences (voltages) around any closed loop or circuit is zero. Mathematically:
Now, for the second part of the question, we use the principle of the metre bridge. The balance condition for a metre bridge is given by:
Where:
- and are the resistances in the bridge,
- is the length on one side of the bridge,
- is the length on the other side. Given: - The original balance point occurs at and . Thus, the equation becomes:
Now, when a resistance of 12 ohms is connected in parallel to , the balance point shifts to a new position. The equivalent resistance of the parallel combination of and is:
The new balance condition is:
Given the new lengths: - and , the ratio becomes:
Now, we solve for : Substituting the expression for : Now, substitute from the earlier equation:
Simplifying the equation, we solve for and find the values of and .

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.
- Kirchhoff's Voltage Law (KVL): The sum of the electrical potential differences (voltages) around any closed loop or circuit is zero.

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:
The current in the bridge is divided between two parallel branches. The current is divided into and at point , where:

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 . Therefore, the currents and are: where 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 and can be calculated using the above equations based on the resistances in the circuit.

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge circuit. The principle of the Wheatstone bridge states that when the bridge is balanced, the ratio of resistances in one leg equals the ratio of resistances in the other leg, which is derived from Kirchhoff's law. The balance condition of the Wheatstone bridge is given by: Where: - are the resistors in the four arms of the bridge,
- is the galvanometer, which detects the balance condition by indicating zero deflection when the bridge is balanced.
In the given circuit: We need to find the value of (which is unknown). Since there is no deflection in the galvanometer, the bridge is balanced. Therefore: Solving for : Now, using Kirchhoff's law, we can find the current flowing through the cell. The total resistance in the circuit is the sum of the resistances in series. The total resistance is: The current flowing through the cell is given by Ohm's law: Thus, the current flowing through the cell is 0.686 A.