In the circuit diagram below, E is a battery of negligible internal resistance. If its...
PHYSICS
WAEC 2010
In the circuit diagram below, E is a battery of negligible internal resistance. If its emf is 9.0V. Calculate the current in the circuit
- A. 1.8A
- B. 1.0A
- C. 0.8A
- D. 0.3A
Correct Answer: B. 1.0A
Explanation
\(\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2}\)
= \(\frac{1}{5} + \frac{1}{20}\)
= \(\frac{4 + 1}{20} = \frac{5}{20}\)
r = \(\frac{20}{5}\)
= 4\(\Omega\)
R = R1 + R2
= 5 + 4
= 9\(\Omega\)
but V = IR
I = \(\frac{V}{R} = \frac{9}{9}\)
= 1A
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