In the circuit diagram above, calculate the current in the 12 Ω resistor if the cell has an emf of 12V and an internal resistance of 1Ω
A.
0.8A
B.
1.0A
C.
1.6A
D.
2.4A
Correct answer is D
\(E = I(R + r)\)
\(I = \frac{E}{R + r}\)
\(\frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}\)
\(R = 4\Omega\)
\(I = \frac{12}{1 + 4} = \frac{12}{5}\)
= 2.4A