JEE MAIN - Physics (2023 - 6th April Evening Shift - No. 9)
Figure shows a part of an electric circuit. The potentials at points $$a, b$$ and $$c$$ are $$30 \mathrm{~V}, 12 \mathrm{~V}$$ and $$2 \mathrm{~V}$$ respectively. The current through the $$20 ~\Omega$$ resistor will be,
_6th_April_Evening_Shift_en_9_1.png)
0.2 A
0.6 A
0.4 A
1.0 A
Explanation
Sum of current at junction point will be zero :
_6th_April_Evening_Shift_en_9_2.png)
$$ \begin{aligned} & \frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0 \\\\ & \Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30} \\\\ & \Rightarrow x\left(\frac{6+3+2}{60}\right)=\frac{180+36+4}{60} \\\\ & \Rightarrow x=\frac{220}{11}=20 \mathrm{~V} \end{aligned} $$
$\therefore$ Current through $20 \Omega=\frac{x-12}{20}$
$$ =\frac{20-12}{20}=\frac{2}{5}=0.4 \mathrm{~A} $$
$$ \begin{aligned} & \frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0 \\\\ & \Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30} \\\\ & \Rightarrow x\left(\frac{6+3+2}{60}\right)=\frac{180+36+4}{60} \\\\ & \Rightarrow x=\frac{220}{11}=20 \mathrm{~V} \end{aligned} $$
$\therefore$ Current through $20 \Omega=\frac{x-12}{20}$
$$ =\frac{20-12}{20}=\frac{2}{5}=0.4 \mathrm{~A} $$
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