Question

4. In the figure below, a current of 2.8A flows

towards the bridge circuit and no current flows
in the galvanometer, G. Calculate the current
flowing in the 4Ω resistor. .
Α. 2.8A Β. 2.4A C. 2.0AD 1.2A
​

Answer 1

The current flowing through the 4Ω resistor is;

C. 2.0 A

Step-by-step explanation:

In the given figure, we have;

The series resistors in the bridge = 8Ω, and 4Ω are in series and 20Ω and 10Ω are in series

The sum of each of the series resistors are;

R₁ = 8Ω + 4Ω = 12Ω

R₂ = 20Ω + 10Ω = 30Ω

The equivalent resistors of the series resistors, R₁ and R₂ are parallel to each other

The current flowing through the 4Ω resistor = The current in R₁ = I₁

By the current divider rule, we have;

`I_1 = (R_2)/(R_1 + R_2) \cdot I`

`I_1 = (30\Omega )/(12 \Omega + 30 \Omega) * 2.8 \, A = 2.0 \, A`

Therefore, I₁ = 2.0 A

The current flowing through the 4Ω resistor = I₁ = 2 A