Answer:
The current flowing through the 4Ω resistor is;
C. 2.0 A
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;
[tex]I_1 = \dfrac{R_2}{R_1 + R_2} \cdot I[/tex]
[tex]I_1 = \dfrac{30\Omega }{12 \Omega + 30 \Omega} \times 2.8 \, A = 2.0 \, A[/tex]
Therefore, I₁ = 2.0 A
The current flowing through the 4Ω resistor = I₁ = 2 A
