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
\(E = I(R + r)\)
\(I = \frac{E}{R + r}\)
\(\frac{1}{R} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}\)
\(R = 4\Omega\)
\(I = \frac{12}{1 + 4} = \frac{12}{5}\)
= 2.4A
The current through the 12 Ω resistor (in parallel) is: = 2.4 × (\(\frac{6}{18}\)) = 2.4 × (\(\frac{1}{3}\)) = 0.8 A
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