Question 255: Current Electricity

Question

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,

Accepted Answer

Sum of current at junction point will be zero :
$$
\begin{aligned}
& \frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0 \\
& \Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}ight)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30} \\
& \Rightarrow x\left(\frac{6+3+2}{60}ight)=\frac{180+36+4}{60} \\
& \Rightarrow x=\frac{220}{11}=20 \mathrm{~V}
\end{aligned}
$$
$	herefore$ Current through $20 \Omega=\frac{x-12}{20}$
$$
=\frac{20-12}{20}=\frac{2}{5}=0.4 \mathrm{~A}
$$

Answer

0.2 A

Answer

0.6 A

Answer

0.4 A

Answer

1.0 A

Current Electricity

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,

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