A portable radio is connected to a 9.000 V battery and draws a current of 0.025 A. What is the resistance of the radio?
Using Ohm’s law, the resistance is $R = \dfrac{V}{I} = \dfrac{9.000\ \text{V}}{0.025\ \text{A}} = 360\ \Omega$. So the portable radio’s resistance is $360\ \Omega$.
What the problem is asking
You are given the voltage of the battery and the current the radio draws. Resistance tells you how much the device opposes the flow of current for that applied voltage.
Use Ohm’s law to relate voltage, current, and resistance
Ohm’s law is $$V = IR$$ Solve for resistance $R$: $$R = \frac{V}{I}$$
Substitute the given values
$$R = \frac{9.000\ \text{V}}{0.025\ \text{A}} = 360\ \Omega$$
Quick reasonableness check
A $9\ \text{V}$ source with a few hundred ohms would give a few hundredths of an amp, since $I = V/R = 9/360 = 0.025\ \text{A}$. That matches the given current.
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