A cell of e.m.f 2V and internal resistance 1\(\Omega\) supplies a current of 0.5 amps to a resistance whose value is
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option E
Explanation:
1 = \(\frac{E}{R + r}\)
\(\frac{2}{R + 1}\) = 0.5
0.5R = 2 - 0.5 = 1.5
R = \(\frac{1.5}{0.5}\)
= 3\(\Omega\)
1 = \(\frac{E}{R + r}\)
\(\frac{2}{R + 1}\) = 0.5
0.5R = 2 - 0.5 = 1.5
R = \(\frac{1.5}{0.5}\)
= 3\(\Omega\)