Question

Determine the current in the 3 Ω branch of a Wheatstone Bridge in the circuit shown in the figure.

Answer 1

From the figure:

Left upper arm = 20 Ω

Left lower arm = 12 Ω

Right upper arm = 2 Ω

Right lower arm = 1 Ω

Central branch = 3 Ω

Checking bridge balance:

`20/12` = 1.67

`2/1` = 2

∴ `20/12 ne 2/1`

∴ The bridge is not perfectly balanced.

However, the 6 V battery is connected across the left and right nodes.

The bridge can be analyzed by simplifying two parallel series branches:

Upper path resistance:

20 + 2 = 22 Ω

Lower path resistance:

12 + 1 = 13 Ω

Let total voltage = 6 V

Voltage division in upper branch: Current in upper branch:

I_u = `6/22`

Voltage drop across 20:

V_A = I_u × 20

= `(6 xx 20)/22`

= 5.45 V

Lower branch current:

I_l = `6/13`

Voltage at lower junction:

V_B = I_l × 12

= `(6 xx 12)/13`

= 5.54 V

Comparing junction potentials:

V_A ≈ 5.45 V, V_B ≈ 5.54 V

It is almost equal. The potential difference across 3 is negligible.

Potential difference across the central branch is nearly zero, so no current flows through the 3 Ω resistor.

∴ The current in the 3 Ω branch of a Wheatstone Bridge in the circuit shown in the figure is 0 A.