Question 6
In exercise 19.4, B1 is 20 Ω in series with B2 (6 Ω) and B3 (12 Ω) in parallel, across 12 V. If B2 burns out, what current flows?
The right answer is the one marked above.
0.375 A. A burnt-out B2 breaks its branch, so only B3 is left. The circuit becomes 20 + 12 = 32 Ω in series, and I = 12 ÷ 32 = 0.375 A.
The trap is to keep treating it as a parallel pair with one value changed. A broken branch is removed from the circuit entirely — redraw before calculating.