WAEC - Physics (2014 - No. 41)

In the circuit diagram above, E is a battery of negligible internal resistance. If its emf is 9.0V. Calculate the current in the circuit ``
1.8A
1.0A
0.8A
0.3A

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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