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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Current in Conductors

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current in Conductors

Determine the current in the 20 resistor for the circuit shown in...

Determine the current in the 20 Ω20 \text{ }\Omega resistor for the circuit shown in the figure.
Determine the current in the 20 resistor for the circuit shown in the figure.

Options

  1. A0.1 A0.1 \text{ A}
  2. B0.2 A0.2 \text{ A}
  3. C0.4 A0.4 \text{ A}
  4. D0.5 A0.5 \text{ A}

Correct answer

B. 0.2 A0.2 \text{ A}

Step-by-step solution

Taking the potential at node B as 0 V0\text{ V} and at node A as VAV_A.

Applying Kirchhoff's Current Law (KCL) at node A:

VA−510+VA−020+VA−510=0\dfrac{V_A - 5}{10} + \dfrac{V_A - 0}{20} + \dfrac{V_A - 5}{10} = 0

Multiplying the equation by 2020 to clear the denominators:

2(VA−5)+VA+2(VA−5)=02(V_A - 5) + V_A + 2(V_A - 5) = 0

2VA−10+VA+2VA−10=02V_A - 10 + V_A + 2V_A - 10 = 0

5VA=20⇒VA=4 V5V_A = 20 \Rightarrow V_A = 4\text{ V}

The current through the 20 Ω20\text{ }\Omega resistor is:

I=VA−VB20=4−020=0.2 AI = \dfrac{V_A - V_B}{20} = \dfrac{4 - 0}{20} = 0.2\text{ A}
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