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QJEEM23Phc701: JEE Main 2023 Physics - Class 12 JEEM23Phc701 Solution: Figure shows a part of an electric circuit. The potentials at…

Figure shows a part of an electric circuit. The potentials at points a,ba, b and cc are 30 V,12 V30 \mathrm{~V}, 12 \mathrm{~V} and 2 V2 \mathrm{~V} respectively. The current through the 20 Ω20 ~\Omega resistor will be,

JEE Main 2023 (Online) 6th April Evening Shift Physics - Current Electricity Question 62 English

Multiple Choice Options

  • A 0.2 A
  • B 0.6 A
  • C 0.4 A
  • D 1.0 A

Step-by-Step Verified Solution

Sum of current at junction point will be zero :

JEE Main 2023 (Online) 6th April Evening Shift Physics - Current Electricity Question 62 English Explanation
amp;x−3010+x−1220+x−230=0amp;⇒x(110+120+130)=3010+1220+230amp;⇒x(6+3+260)=180+36+460amp;⇒x=22011=20 V\begin{aligned}& \frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0 \\\\& \Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30} \\\\& \Rightarrow x\left(\frac{6+3+2}{60}\right)=\frac{180+36+4}{60} \\\\& \Rightarrow x=\frac{220}{11}=20 \mathrm{~V}\end{aligned}

∴\therefore Current through 20Ω=x−122020 \Omega=\frac{x-12}{20}

=20−1220=25=0.4 A=\frac{20-12}{20}=\frac{2}{5}=0.4 \mathrm{~A}

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