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
- 0.8A
- 1.0A
- 1.6A
-
2.4A
The correct answer is: D
Explanation
\(E = I(R + r)\)
\(I = \frac{E}{R + r}\)
\(\frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}\)
\(R = 4\Omega\)
\(I = \frac{12}{1 + 4} = \frac{12}{5}\)
= 2.4A