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
The correct answer is: B
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