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

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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 = E²R/(R + r)². For 3 Ω, P₁ = 100 × 3/5² = 12 W. For 8 Ω, P₂ = 100 × 8/10² = 8 W. Thus P₁:P₂ = 12:8 = 3:2.

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

Concept context

More Current Electricity questions

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?