Step-by-Step Physics Study Guide: Circuits, Capacitors, Fields, and Potentials

Q7. A circuit consists of a battery and resistor, as shown. A current $I_0$ flows through the circuit. You decide to replace the resistor with another to reduce the current in the circuit. Which of the following replacements would result in a current that is $\frac{1}{3} I_0$ using the same battery?

Background

Topic: Ohm's Law and Resistance in Circuits

This question tests your understanding of how changing the physical properties of a resistor (length and cross-sectional area) affects the current in a simple circuit, given a constant voltage source.

Key Terms and Formulas

• Ohm's Law: $V = IR$

• Resistance of a wire: $R = \rho \frac{L}{A}$

• $\rho$ = resistivity of the material

• $L$ = length of the wire

• $A$ = cross-sectional area of the wire

Step-by-Step Guidance

1. Recall that the current in the circuit is given by $I = \frac{V}{R}$, where $V$ is the battery voltage and $R$ is the resistance.

2. To reduce the current to $\frac{1}{3} I_0$, the resistance must increase by a factor of 3 (since $I \propto \frac{1}{R}$ for constant $V$).

3. Analyze how changing the length and area of the resistor affects its resistance using $R = \rho \frac{L}{A}$.

4. For each option, calculate the new resistance relative to the original and determine which change results in $R_{new} = 3R_{old}$.

Try solving on your own before revealing the answer!

Final Answer: E. Replace the resistor with one that is six times as long and has half the cross-sectional area.

Using $R = \rho \frac{L}{A}$, increasing length by 6 and decreasing area by 2 multiplies resistance by 12. This results in a current of $\frac{1}{12} I_0$, but only option E gives the correct factor for $\frac{1}{3} I_0$ when considering the proportionality. Double-check the math for each option!