ISRO EC 2023 Q.30: solution and answer

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

1. R1 and R2 are in series with the parallel pair R3 and R4, and the whole network is across the source V. All the source current flows through R1, so $I_{R_1}=\frac{V}{R_1+R_2+(R_3\parallel R_4)}$.

2. V is fixed, so the current is largest when the total resistance is smallest and smallest when the total resistance is largest: $\frac{I_{\max }}{I_{\min }}=\frac{R_{T,\max }}{R_{T,\min }}$.

3. The four values $3,6,10,15\Omega$ are assigned one to each resistor, so the only choice is which two form the parallel pair; the other two are in series. Using $R_T=(\text{series pair sum})+\frac{ab}{a+b}$:

4. Parallel $\{3,6\}$: $2\Omega$, series $10+15=25\Omega$, total $27\Omega$. Parallel $\{3,10\}$: $2.31\Omega$, series $6+15=21\Omega$, total $23.31\Omega$. Parallel $\{3,15\}$: $2.5\Omega$, series $6+10=16\Omega$, total $18.5\Omega$.

5. Parallel $\{6,10\}$: $3.75\Omega$, series $3+15=18\Omega$, total $21.75\Omega$. Parallel $\{6,15\}$: $4.29\Omega$, series $3+10=13\Omega$, total $17.29\Omega$. Parallel $\{10,15\}$: $6\Omega$, series $3+6=9\Omega$, total $15\Omega$.

6. Hence $R_{T,\max }=27\Omega$ and $R_{T,\min }=15\Omega$, and $$\frac{I_{\max }}{I_{\min }}=\frac{27}{15}=1.8$$