A Wheatstone bridge has four resistances 200Ω, 20Ω, 400Ω, and 40Ω. If the bridge is connected to a 1.5 V battery, calculate the currents through individual resistors.

First arm resistance P=200Ω

Second arm resistance Q=20Ω

Third arm resistance R=400Ω

Fourth arm resistance S=40Ω

The potential difference is VAC =1.5V since points A and C are connected to the battery.

Therefore, the ratios of the arms,

P/Q=200/20 = 10

R/S= 400/40 = 10

Hence it satisfies the null condition that is, P/Q = R/S

∴ Current in the resistance P = current in the resistance VAB/(P+Q)

That is, 1.5/ (200+20) = 0.0681A

Again, current in the resistance R = current in the resistance VAB/(R+S)

That is, 1.5/ (400+40) = 0.0340A