SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Given, R = 16 Ω
Therefore, the circuit becomes Step 1: Solving 16 Ω and 4 Ω in series combination
16 Ω + 4 Ω = 20 Ω Step 2: Solving both 20 Ω in parallel combination
$\mathit{R}=\frac{20\times 20}{20+20}=10\mathrm{\Omega }$ Step 3: Here 10 Ω and 2 Ω is connected in series
$\therefore {\mathrm{R}}_{\mathrm{eq}}=12\mathrm{\Omega }$
Hence, Ohmmeter will be measured Resistance of 12 Ω