Question

calculate the potential difference between the junctions B and D in the wheatstone bridge shown in figure​

Answer 1

Given :

Voltage applied in the circuit = 2 V

Resistance between junction A and B = 1 ohm

Resistance between  junction B and C = 1 ohm

Resistance between  junction C and D = 1 ohm

Resistance between  junction A and D = 1.5 ohm

To Find :

The magnitude of potential difference between the junctions B and D = ?

Solution :

Since by Ohm's Law we have :

V = IR              - (1)

Since eqv resistance in branch ABC is 2 ohm , so the current $I_1$ is :

=$\frac{2}{2}$

= 1 A

And in the branch ADC eqv resistance is 2.5 ohm , so current $I_2$ is :

= $\frac{2}{2.5}$

= 0.8 A

Now , $V_B = V_A - 1 \times I_1$        ( ∴resistance is 1 ohm )

Or , $V_B = V_A - 1$              - (2)

And , $V_D = V_A - 1.5 \times I_2$

Or, $V_D = V_A - 1.2$            - (3)

So , the potential difference between B and D is :

=$V_B - V_D$

=$(V_A - 1 ) -(V_A - 1.5 )$

= -1 + 1.2

= 0.2 V

Hence , the potential difference between the junctions B and D is 0.2 V .