The current I in the fig above is
Attached Figure / Diagram
A
4.00A
B
1.30A
C
0.80A
D
0.75A
Verified Correct Answer: A.
Physical Principles & Numerical Solution
Total resistance in series Ra = 4\(\Omega\) and another 4 \(\Omega\)
Forming two resistors in parallel: \(\frac{R_a \times R_b}{R_a + R_b}\) = \(\frac{4 \times 4}{4 + 4}\) = 2 (\Omega\)
and this is in series with the 1 (\Omega\) resistor = 3(\Omega\)
I = \(\frac{12}{3}\) = 4.0A
Source: Myschool Past Questions Archive (Record #41719)