In a series–parallel circuit consisting of R₁ and R₂ in series and connected in parallel with R₃, if R₁ opens, then R₂'s voltage will:

ANS:B - decrease

If we take r1=2, r2=3 & r3=4.

Then,

r1+r2 = 5 //With r3 find value total resistance.

RT = 2.22.

Image current flow is 1 amp.

Then voltage is V = ir = 1*2.22 = 2.22 v.

If r1 is open, then find voltage r2/r3 = 1.714.

Current is 1 amp.

So V = IR 1.714*1 = 1.741 v.

Voltage decrease B is correct.