Question

1. the probable fault in this circuit is

a l2 open

b l1 open

c a component changed value in r2, l2 branch

d a component changed value in r1, l1 branch

Explanation:

Step1: Calculate current in R1,L1 branch if normal

Using Ohm's law \(I = \frac{V}{R}\). For the R1,L1 branch, if only resistance \(R = 1k\Omega=1000\Omega\) and \(V = 60V_{pp}\), the predicted current \(I=\frac{60}{1000}=60mA\) (ignoring inductive reactance for a quick check as in an open - circuit case). But the predicted current is \(12.1mA\). If \(L1\) is open, the current through \(R1\) would be calculated using \(I=\frac{V_{across R1}}{R1}\). Given \(V = 12.1V\) across \(R1\) (\(R1 = 1k\Omega\)), \(I=\frac{12.1}{1000}=12.1mA\)

Step2: Check other options

• If \(L2\) is open: The current through \(R1,L1\) branch would be affected by its own components. But the current in \(R1,L1\) branch is abnormal. If \(L2\) is open, it mainly affects the \(R2,L2\) branch.
• If a component changed value in \(R2,L2\) branch: It would not directly cause the abnormal current in \(R1,L1\) branch as observed (\(12.1mA\) instead of expected higher value considering source voltage).
• If a component changed value in \(R1,L1\) branch (not open): If it was a value change (not open), the voltage across \(R1\) would not be such a small fraction (\(12.1V\) out of \(60V_{pp}\)) in a simple series - like (with \(L1\)) circuit as an open - circuit causes a voltage division based on open (infinite impedance)

Answer:

B. L1 open