Answer:
(D) V > V' > 1/2 V
Step-by-step explanation:
When B₂ is in the circuit, the equivalent resistance is:
Req = R + (1/R + 1/2R)^-1
Req = R + 2/3 R
Req = 5/3 R
So the current in B₁ is:
I = V/R
I = V₀ / (5/3 R)
I = 3/5 V₀/R
Therefore, the reading of V₁ is:
V = IR
V₁ = (3/5 V₀/R) R
V₁ = 3/5 V₀
V₁ = 0.6 V₀
After B₂ is removed, the equivalent resistance is:
Req = 3 R
So the current in B₁ is:
I = V/R
I = V₀ / (3 R)
I = 1/3 V₀/R
Therefore, the new reading V₁' is:
V = IR
V₁' = (1/3 V₀/R) R
V₁' = 1/3 V₀
V₁' = 0.333 V₀
The new voltage is less than the original voltage, but more than half the original voltage. The answer is D.