An 8.00-Ω resistor is connected to a battery with internal resistance of 0.200 Ω.If the terminal voltage across the battery is 9.00 V,what is the battery's emf?
A)8.78 Ω
B)9.23 Ω
C)0.225 Ω
D)0.178 Ω

Question

Quizplus · Accepted Answer

From Kirchhoff's loop rule applied to the circuit:

emf - IR - Ir = 0,

where I is the current in the circuit.

Solving for emf gives:

emf = IR + Ir = (8.00 Ω)(I) + (0.200 Ω)(I) = (8.20 Ω)(I).

The terminal voltage across the battery is given by V = emf - Ir = 9.00 V, so we have:

9.00 V = (8.20 Ω)(I) - (0.200 Ω)(I) = (8.00 Ω)(I).

Solving for I gives:

I = V / (8.00 Ω) = 1.125 A.

Substituting back into the expression for emf gives:

emf = (8.20 Ω)(1.125 A) = 9.23 V.

Therefore, the correct answer is B.

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