Question

Establish the relation between the resistances of arms of wheatstone bridge in balance conditions.
                                                                           OR 
The ratio of lengths and masses of three wires of same metal are 3 : 2 : 1 and 1 : 2 : 3 respectively. Find the ratio of resistances of those wires.} 
 

Hint:

For the OR part, the key is to combine the formulas for resistance and mass to find how resistance depends on length and mass. The relation \( R \propto L^2/m \) is very useful for such comparison problems.

Answer 1

Part I: Wheatstone Bridge 
Step 1: Principle and Diagram: 
A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit. The circuit consists of four resistors P, Q, R, and S arranged in a quadrilateral shape. A galvanometer G is connected between two opposite junctions, and a voltage source is connected to the other two junctions. 
 
Step 2: Balanced Condition and Derivation: 
The bridge is said to be balanced when no current flows through the galvanometer (\(I_g = 0\)). This occurs when the potential at point B is equal to the potential at point D (\(V_B = V_D\)). 
Let the current from the source split at A into \(I_1\) (through P) and \(I_2\) (through R). 
When the bridge is balanced (\(I_g = 0\)), the current \(I_1\) also flows through Q, and the current \(I_2\) also flows through S. 
Since \(V_B = V_D\), the potential drop across P must be equal to the potential drop across R. \[ V_A - V_B = V_A - V_D \implies I_1 P = I_2 R \quad .s (1) \] Similarly, the potential drop across Q must be equal to the potential drop across S. \[ V_B - V_C = V_D - V_C \implies I_1 Q = I_2 S \quad .s (2) \] Dividing equation (1) by equation (2), we get: \[ \frac{I_1 P}{I_1 Q} = \frac{I_2 R}{I_2 S} \] \[ \frac{P}{Q} = \frac{R}{S} \] This is the required relation for a balanced Wheatstone bridge. 
Part II: OR Question 
Step 1: Key Formulas: 
The resistance R of a wire is given by \( R = \rho \frac{L}{A} \), where \(\rho\) is the resistivity, L is the length, and A is the cross-sectional area. 
The mass m of a wire is given by \( m = \text{Volume} \times \text{Density} = (A . L) . d \). 
Step 2: Deriving the Resistance Ratio Formula: 
From the mass formula, we can express the area A as \( A = \frac{m}{Ld} \). 
Substituting this into the resistance formula: \[ R = \rho \frac{L}{A} = \rho \frac{L}{(m/Ld)} = \frac{\rho d L^2}{m} \] Since the three wires are made of the "same metal", their resistivity (\(\rho\)) and density (\(d\)) are the same. Therefore, the resistance is proportional to \(L^2/m\). \[ R \propto \frac{L^2}{m} \] So, the ratio of their resistances will be: \[ R_1 : R_2 : R_3 = \frac{L_1^2}{m_1} : \frac{L_2^2}{m_2} : \frac{L_3^2}{m_3} \] Step 3: Calculation: 
We are given the ratios:
Ratio of lengths, \( L_1 : L_2 : L_3 = 3 : 2 : 1 \)
Ratio of masses, \( m_1 : m_2 : m_3 = 1 : 2 : 3 \)
Substituting these ratios into our derived proportion: \[ R_1 : R_2 : R_3 = \frac{(3)^2}{1} : \frac{(2)^2}{2} : \frac{(1)^2}{3} \] \[ R_1 : R_2 : R_3 = \frac{9}{1} : \frac{4}{2} : \frac{1}{3} \] \[ R_1 : R_2 : R_3 = 9 : 2 : \frac{1}{3} \] To express this ratio in integers, we multiply all parts by 3: \[ R_1 : R_2 : R_3 = 9 \times 3 : 2 \times 3 : \frac{1}{3} \times 3 \] \[ R_1 : R_2 : R_3 = 27 : 6 : 1 \] Step 4: Final Answer: 
The ratio of the resistances of the three wires is 27 : 6 : 1.