JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2024
- Hard
- Numerical
The current flowing through the \(1 \Omega\) resistor is \(\frac{\mathrm{n}}{10} \ A.\) The value of \(n\) is _______.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(\frac{y-5}{2}+\frac{y-0}{2}+\frac{y-x+10}{1}=0\)
\(y-5+y+2 y-2 x+20=0\)
\(4 y-2 x+15=0 \quad \ldots .(i)\)
\(\frac{x-5}{4}+\frac{x-0}{4}+\frac{x-10-y}{1}=0\)
\(x-5+x+4 x-40-4 y=0\)
\(6 x-4 y-45=0 \quad . .(i)\)
\(\frac{-2 x+4 y+15=0 . .(i i)}{4 x-30=0}\)
\(x=\frac{15}{2} \& 4 y-15+15=0\)
\(y=0\)
\(i=\frac{y-x+10}{1}\)
\(i=\frac{0-7.5+10}{1}\)
\(i=2.5 A=\frac{n}{10} A\)
\(n=25\)

\(y-5+y+2 y-2 x+20=0\)
\(4 y-2 x+15=0 \quad \ldots .(i)\)
\(\frac{x-5}{4}+\frac{x-0}{4}+\frac{x-10-y}{1}=0\)
\(x-5+x+4 x-40-4 y=0\)
\(6 x-4 y-45=0 \quad . .(i)\)
\(\frac{-2 x+4 y+15=0 . .(i i)}{4 x-30=0}\)
\(x=\frac{15}{2} \& 4 y-15+15=0\)
\(y=0\)
\(i=\frac{y-x+10}{1}\)
\(i=\frac{0-7.5+10}{1}\)
\(i=2.5 A=\frac{n}{10} A\)
\(n=25\)

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