GATE 2020 IN Q.4: solution and answer

Official answer: CSource: Official answer key, GATE 2020 IN (IIT Delhi)

1. One loop, one current. The circuit has a single loop, so the $100\Omega$, $2\Omega$ and $1\Omega$ resistors are in series and all carry the same current $I$ (KCL at every node).

2. Current limit of each resistor. A resistor dissipates $P=I^{2}R$, so the largest current it can carry without exceeding its rated power is $I_{\max }=\sqrt{P_{\text{rated}}/R}$.

3. $100\Omega ,1\text{W}$: $I_{\max }=\sqrt{\frac{1}{100}}=0.1\text{A}$

4. $2\Omega ,0.5\text{W}$: $I_{\max }=\sqrt{\frac{0.5}{2}}=\sqrt{0.25}=0.5\text{A}$

5. $1\Omega ,0.25\text{W}$: $I_{\max }=\sqrt{\frac{0.25}{1}}=0.5\text{A}$

6. The weakest element sets the limit. Every resistor must stay within its rating, so the common series current can be no larger than the smallest of these limits: $I_{\text{safe}}=\min (0.1,0.5,0.5)=0.1\text{A}$.

7. Check. At $I=0.1\text{A}$: $P_{100}=(0.1{)}^2\times 100=1\text{W}$ (exactly at its rating), $P_2=(0.1{)}^2\times 2=0.02\text{W}$ and $P_1=(0.1{)}^2\times 1=0.01\text{W}$, both well below their ratings. The answer is (C).