a chemical cell of internal resistance 1\(\Omega\) supplies electric current to an external resistor of resistance 3\(\Omega\). Calculate the efficiency of the cell
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Correct Answer: Option A
Explanation:
R = \(r_1 + R = 1 + 3 = 4\Omega\)
\(\frac{3}{4} \times \frac{100}{1} = 75\)%
R = \(r_1 + R = 1 + 3 = 4\Omega\)
\(\frac{3}{4} \times \frac{100}{1} = 75\)%