A cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of...
PHYSICS
WAEC 2006
A cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of resistance 2.0\(\Omega\) and 3.0\(\Omega\) in series. Calculate the current through the resistors
- A. 0.25A
- B. 0.30A
- C. 0.35A
- D. 0.50A
Correct Answer: A. 0.25A
Explanation
I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)
= \(\frac{1.5}{6} = 0.25A\)
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