Step-by-Step Guidance for Circuit Analysis: Capacitors and Resistors

Q7. For the capacitor circuit below, determine the charge, voltage, and energy stored on each capacitor:

Background

Topic: Capacitors in Series and Parallel Circuits

This question tests your understanding of how to analyze a circuit containing capacitors in both series and parallel arrangements. You will need to find the equivalent capacitance, the voltage across each capacitor, the charge stored, and the energy stored in each capacitor.

Key Terms and Formulas

• Capacitance in series: $\frac{1}{C_{eq}} = \frac{1}{C_2} + \frac{1}{C_3}$

• Capacitance in parallel: $C_{eq} = C_1 + C_{23}$

• Charge on a capacitor: $Q = C \cdot V$

• Energy stored in a capacitor: $U = \frac{1}{2} C V^2$

Step-by-Step Guidance

1. First, identify which capacitors are in series and which are in parallel. In this circuit, $C_2$ and $C_3$ are in series, and their combination is in parallel with $C_1$.

2. Calculate the equivalent capacitance of $C_2$ and $C_3$ in series using $\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3}$, where $C_2 = 25\ \mu F$ and $C_3 = 87\ \mu F$.

3. Add $C_1$ (73 $\mu$F) in parallel to the series combination to find the total equivalent capacitance: $C_{eq} = C_1 + C_{23}$.

4. Find the total charge supplied by the battery using $Q_{total} = C_{eq} \cdot V_{battery}$, where $V_{battery} = 4.5\ V$.

5. Determine the voltage across $C_1$ (it is in parallel with the battery, so $V_{C_1} = V_{battery}$) and the voltage across the series combination ($C_2$ and $C_3$ share the remaining voltage).

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Q8. For the resistor circuit below, determine the current, voltage, and power of each resistor:

Background

Topic: Series and Parallel Resistor Circuits

This question tests your ability to analyze a circuit with resistors in both series and parallel. You will need to use Ohm's Law and the rules for combining resistors to find the current through, voltage across, and power dissipated by each resistor.

Key Terms and Formulas

• Ohm's Law: $V = IR$

• Resistors in series: $R_{eq} = R_1 + R_2 + ...$

• Resistors in parallel: $\frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_3}$

• Power dissipated: $P = I^2 R$ or $P = V^2 / R$

Step-by-Step Guidance

1. Identify which resistors are in series and which are in parallel. In this circuit, $R_2$ and $R_3$ are in parallel, and their combination is in series with $R_1$.

2. Calculate the equivalent resistance of $R_2$ and $R_3$ in parallel using $\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3}$, where $R_2 = 33\ \Omega$ and $R_3 = 62\ \Omega$.

3. Add $R_1$ (75 $\Omega$) in series to the parallel combination to find the total equivalent resistance: $R_{eq} = R_1 + R_{23}$.

4. Find the total current from the battery using $I_{total} = \frac{V_{battery}}{R_{eq}}$, where $V_{battery} = 4.5\ V$.

5. Determine the voltage across $R_1$ (using $V = IR$) and the voltage across the parallel branch (it will be the same for both $R_2$ and $R_3$).

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Q10. For the circuit below, determine currents I1, I2, and I3:

Background

Topic: Kirchhoff's Rules for Multi-Loop Circuits

This question tests your ability to apply Kirchhoff's Junction Rule and Loop Rule to solve for unknown currents in a multi-loop circuit.

Key Terms and Formulas

• Junction Rule: $\sum I_{in} = \sum I_{out}$

• Loop Rule: $\sum V = 0$ around any closed loop

• Ohm's Law: $V = IR$

Step-by-Step Guidance

1. Assign current directions for $I_1$, $I_2$, and $I_3$ as shown in the diagram.

2. Apply the Junction Rule at the node where the three resistors meet: $I_1 + I_2 = I_3$ (or similar, depending on your current directions).

3. Write a loop equation for the left loop (including the 7.8 V battery, 40 $\Omega$ resistor, and 54 $\Omega$ resistor) using the Loop Rule.

4. Write a loop equation for the right loop (including the 23.9 V battery, 59 $\Omega$ resistor, and 54 $\Omega$ resistor) using the Loop Rule.

5. Set up the system of equations and prepare to solve for the unknown currents. (Stop here before solving!)

Try solving on your own before revealing the answer!