Four identical light bulbs (b1, b2, b3, and b4) and
six voltmeters are connected as shown in the figure.
assuming that v is as its original reading when all bulbs
are working and v' is as its reading when b2 is caught
on fire. if b2 were to be caught on fire, the reading of v
is

(D) V > V' > 1/2 V

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.

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Explanation: