Physics 1201: Midterm I Review – Circuits, Capacitors, and Electric Forces

Q8. Find the current flowing in the circuit below if $\varepsilon_1 = 12\ \mathrm{V}$, $\varepsilon_2 = 9.0\ \mathrm{V}$, $R_1 = 100\ \Omega$ and $R_2 = 160\ \Omega$.

Background

Topic: DC Circuits – Kirchhoff's Loop Rule

This question tests your ability to analyze a simple series circuit with two batteries and two resistors using Kirchhoff's loop rule to find the current.

Key Terms and Formulas

• Kirchhoff's Loop Rule: The sum of the potential differences around any closed loop in a circuit is zero.

• Ohm's Law: $V = IR$

For this circuit, the loop equation is:

$\varepsilon_1 - I R_1 - \varepsilon_2 - I R_2 = 0$

Step-by-Step Guidance

1. Write the loop equation for the circuit, starting at the negative terminal of $\varepsilon_1$ and moving clockwise:

   $\varepsilon_1 - I R_1 - \varepsilon_2 - I R_2 = 0$

2. Combine like terms to group all terms involving $I$ together:

   $\varepsilon_1 - \varepsilon_2 = I(R_1 + R_2)$

3. Rearrange the equation to solve for the current $I$:

   $I = \frac{\varepsilon_1 - \varepsilon_2}{R_1 + R_2}$

4. Substitute the given values: $\varepsilon_1 = 12\ \mathrm{V}$, $\varepsilon_2 = 9.0\ \mathrm{V}$, $R_1 = 100\ \Omega$, $R_2 = 160\ \Omega$.

Try solving on your own before revealing the answer!

Final Answer: $I = 0.012\ \mathrm{A} = 12\ \mathrm{mA}$

Using $I = \frac{12 - 9}{100 + 160}$, you find the current is $0.012\ \mathrm{A}$ or $12\ \mathrm{mA}$.

This is the current flowing through the circuit, determined by the net emf and total resistance.

Q9. Which one of the following equations is not correct for the circuit shown?

Background

Topic: Kirchhoff's Rules for Multi-Loop Circuits

This question tests your understanding of how to apply Kirchhoff's loop and junction rules to a complex circuit with multiple loops and branches.

Key Terms and Formulas

• Kirchhoff's Junction Rule: The sum of currents entering a junction equals the sum of currents leaving.

• Kirchhoff's Loop Rule: The sum of the potential differences around any closed loop is zero.

Step-by-Step Guidance

1. Examine each equation and identify whether it correctly applies the loop or junction rule to the circuit shown.

2. For loop equations, check that the sum of voltage drops and rises around the loop equals zero.

3. For junction equations, check that the sum of currents entering and leaving a node is balanced.

4. Compare each equation to the actual circuit layout and current directions to see if any equation violates the rules.

Try solving on your own before revealing the answer!

Final Answer: b) $I_3 R_3 + I_4 R_4 = 0$ is not correct.

This equation suggests that the voltage drops across $R_3$ and $R_4$ sum to zero, which is not generally true unless both currents are zero.

Q10. Consider two circuits below, where $C_1 = 2.0\ \mathrm{mF}$, $C_2 = 3.0\ \mathrm{mF}$, $C_3 = 2.0\ \mathrm{mF}$, and $C_4 = 3.0\ \mathrm{mF}$. Assume that $V_1$ is the potential difference across $C_1$, and $Q_1$ is the charge on $C_1$, etc. for the other capacitors. Which one of the following statements is true?

Background

Topic: Capacitors in Series and Parallel

This question tests your understanding of how voltage and charge are distributed across capacitors in series and parallel arrangements.

Key Terms and Formulas

• Series Capacitors: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$; charge $Q$ is the same on each capacitor, voltage divides.

• Parallel Capacitors: $C_{eq} = C_3 + C_4$; voltage $V$ is the same across each, charge divides.

• Charge on a Capacitor: $Q = CV$

Step-by-Step Guidance

1. For the series circuit (left), recall that the charge on each capacitor is the same, but the voltage divides according to capacitance.

2. For the parallel circuit (right), recall that the voltage across each capacitor is the same, but the charge divides according to capacitance.

3. Review each statement and determine if it matches the rules for series or parallel capacitors.

4. Identify which statement correctly describes the relationships for the given circuits.

Try solving on your own before revealing the answer!

Final Answer: b) $Q_1 = Q_2$ and $V_3 = V_4$ is true.

In series, capacitors have equal charge; in parallel, capacitors have equal voltage.