A 12.0 V battery is placed across a 4.00 $\Omega$ resistor.If the current through the resistor is 2.80 A,what is the internal resistance of the battery?
A) 0.200 $\Omega$
B) 0.290 $\Omega$
C) 4.29 $\Omega$
D) 3.71 $\Omega$
E) 0.800 $\Omega$

Question

Quizplus · Accepted Answer

The total voltage is 12.0 V, and the voltage across the 4.00 $\varOmega$ resistor is $2.80 \, \text{A} \times 4.00 \, \varOmega = 11.2 \, \text{V}$. The voltage drop due to internal resistance is $12.0 \, \text{V} - 11.2 \, \text{V} = 0.8 \, \text{V}$. Using $V = IR$, the internal resistance $r$ is $0.8 \, \text{V} / 2.80 \, \text{A} = 0.286 \, \varOmega$, which rounds to 0.290 $\varOmega$.

A 120 V Battery Is Placed Across a 4 Ω\OmegaΩ ResistorIf the Current Through the Resistor Is 2

A 120 V Battery Is Placed Across a 4 $\Omega$ ResistorIf the Current Through the Resistor Is 2