Use the diagram above to calculate the current, I.
A.
0.60A
B.
0.97A
C.
1.03A
D.
5.00A
Correct answer is B
\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)
= \(\frac{1}{6} + \frac{1}{4}\)
R = \(\frac{12}{5} = 2.5\Omega\)
R = 2.4 + 10 = 12.4
V = IR
I = \(\frac{v}{R} = \frac{12}{12.4} = 0.97\)