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⚛️ Physics  ·  Current Electricity  ·  NEET & JEE

Two cells of emf 10 V and 4 V, each having internal resistance 2 Ω, are connected to a common 2 Ω resistor in two adjacent loops. The cells drive currents clockwise in their respective loops. What is the current through the common resistor?

Answer: 1.0 A.

  • A 0.5 A
  • B 2.0 A
  • C 1.0 A
  • D 3.0 A

Correct answer: C. 1.0 A

Explanation: Let the clockwise loop currents be I<sub>1</sub> and I<sub>2</sub>. Kirchhoff equations are 10 = 2I<sub>1</sub> + 2(I<sub>1</sub> - I<sub>2</sub>) and 4 = 2I<sub>2</sub> + 2(I<sub>2</sub> - I<sub>1</sub>). Solving gives I<sub>1</sub> = 4 A and I<sub>2</sub> = 3 A. The common resistor carries |I<sub>1</sub> - I<sub>2</sub>| = 1 A.

SeriesbatteryR1R2I (same in R1, R2)ParallelbatteryR1R2I1I2

In series, the same current flows through both resistors; in parallel, the current splits between branches and the voltage across each resistor is the same.

Concept context

Ohm's law, resistors, Kirchhoff's laws, Wheatstone bridge, and cells.

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