In the circuit diagram above, E is a battery of negligible internal resistance. If its…

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 “

Correct Answer: Option 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