Three electric cells each of e.m.f 1.5V and internal resistance of 1.0\(\Omega\) are connected in parallel across an external resistance of \(\frac{2}{3}\)\(\Omega\). Calculate the value of the current in the resistor
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
E.m.f = 1.5
r = \(\frac{1 \times\ 1 \times 1}{(1 \times\ 1) + (1 \times\ 1) + (1 \times\ 1)}\)
r = \(\frac{1}{3}\)\(\Omega\)
I = \(\frac{E.m.f}{r + R}\)
I = \(\frac{1.5}{\frac{1}{3}\ + \frac{2}{3}}\)
I = 1.5A
E.m.f = 1.5
r = \(\frac{1 \times\ 1 \times 1}{(1 \times\ 1) + (1 \times\ 1) + (1 \times\ 1)}\)
r = \(\frac{1}{3}\)\(\Omega\)
I = \(\frac{E.m.f}{r + R}\)
I = \(\frac{1.5}{\frac{1}{3}\ + \frac{2}{3}}\)
I = 1.5A