JEE Mains · Physics · STD 12 - 3. current electricity
Figure shows a part of an electric circuit. The potentials at points \(a , b\) and \(c\) are \(30\,V , 12\,V\) and \(2\,V\) respectively. The current through the \(20 \Omega\) resistor will be \(........\,A\)
- A \(0.4\)
- B \(0.2\)
- C \(0.6\)
- D \(1.0\)
Answer & Solution
Correct Answer
(A) \(0.4\)
Step-by-step Solution
Detailed explanation
sum of current at junction point will be zero: \(\frac{x-30}{10}+\frac{x-12}{20}+\frac{x-2}{30}=0\) \(\Rightarrow x\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}\right)=\frac{30}{10}+\frac{12}{20}+\frac{2}{30}\) \(\Rightarrow x\left(\frac{6+3+2}{60}\right)=\frac{180+36+4}{60}\)…
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