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⚛️ Physics  ·  Current Electricity  ·  NEET & JEE

A cell of emf 10 V and internal resistance 2 Ω supplies a load. When the load resistance changes from 3 Ω to 8 Ω, what is the ratio of the powers delivered to the load in the two cases?

Answer: 3:2.

  • A 2:3
  • B 4:3
  • C 3:2
  • D 3:4

Correct answer: C. 3:2

Explanation: Power delivered is P = E<sup>2</sup>R/(R + r)<sup>2</sup>. For 3 &Omega;, P<sub>1</sub> = 100 &times; 3/5<sup>2</sup> = 12 W. For 8 &Omega;, P<sub>2</sub> = 100 &times; 8/10<sup>2</sup> = 8 W. Thus P<sub>1</sub>:P<sub>2</sub> = 12:8 = 3:2.

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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