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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Current Electricity

JIPMER 2018 · Physics · Current Electricity

Find i in shown figure

Find i in shown figure Find i in shown figure

Options

  1. A0.2A0.2 A
  2. B0.1A0.1 A
  3. C0.3A0.3 A
  4. D0.4A0.4 A

Correct answer

B. 0.1A0.1 A

Step-by-step solution

∵60Ω\because 60 \Omega and 30Ω30 \Omega resistor are connected in parallel. So, their net resistance Rnet =30×6030+60=180090=20Ω∴i=2Rnet =220=0.1A\begin{aligned} & \mathrm{R}_{\text{net }}=\frac{30 \times 60}{30+60}=\frac{1800}{90}=20 \Omega \\ & \therefore \quad \mathrm{i}=\frac{2}{\mathrm{R}_{\text{net }}}=\frac{2}{20}=0.1 \mathrm{ A} \\ & \end{aligned}
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