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: Option B

B. 1.0A

Explanation

1/r=1/r1+1/r2

= 1/5+1/20

= (4+1)/20=5/20

r = 20/5

= 4Ω

R = R1 + R2

= 5 + 4

= 9Ω

but V = IR

I = V/R=9/9

= 1A