Step-by-Step Guidance for EECE 210 Electric Circuits Quiz Problems

Q1. Calculate the power dissipated by the 6 Ω resistor for R = {2, 22, 32, 62} Ω.

Background

Topic: DC Circuit Analysis, Power Dissipation

This question tests your ability to analyze a circuit using Kirchhoff's Voltage Law (KVL) and calculate the power dissipated in a resistor.

Key Terms and Formulas

• KVL (Kirchhoff's Voltage Law): The sum of voltages around a closed loop is zero.

• Power dissipated in a resistor: $P = I^2 R$

• Current calculation: Use KVL to solve for $I$.

Step-by-Step Guidance

1. Write the KVL equation for the loop: $-20 + RI + 2v_0 + 6I = 0$.

2. Express $v_0$ in terms of $I$: $v_0 = 6I$.

3. Substitute $v_0$ into the KVL equation: $-20 + RI + 12I + 6I = 0$.

4. Simplify to get $(18 + R)I = 20$ and solve for $I$: $I = \frac{20}{R + 18}$.

5. Set up the power formula for the 6 Ω resistor: $P_6 = 6 \left(\frac{20}{R + 18}\right)^2$.

Try solving on your own before revealing the answer!

Final Answer: See below for each R value

Plug in each value of R into the formula $P_6 = 6 \left(\frac{20}{R + 18}\right)^2$ to get the power dissipated for each case.

This approach uses KVL and the power formula to find the answer.

Q2. In the circuit shown, X = {16, 18, 20, 24} Volts. Determine $I_T$.

Background

Topic: Series-Parallel Resistance, Ohm's Law

This question tests your ability to find the total current in a circuit by calculating the equivalent resistance and applying Ohm's Law.

Key Terms and Formulas

• Equivalent resistance ($R_T$): Combine resistors in series and parallel.

• Ohm's Law: $I = \frac{V}{R}$

Step-by-Step Guidance

1. Calculate the equivalent resistance: $R_T = 2 + \frac{4 \times 4}{4 + 4} + 2 + 2 = 8 \Omega$.

2. Use Ohm's Law to set up the current formula: $I_T = \frac{X}{8}$.

Try solving on your own before revealing the answer!

Final Answer: $I_T = \frac{X}{8}$ A

Substitute each value of X to find the total current for each case.

Q3. Find the current $i_o$ in the circuit for R = {2, 4, 7, 12} kΩ.

Background

Topic: Mesh/Supermesh Analysis

This question tests your ability to use mesh analysis and dependent sources to solve for current in a circuit.

Key Terms and Formulas

• Mesh analysis: Apply KVL to loops in the circuit.

• Dependent source: A source whose value depends on a circuit variable.

Step-by-Step Guidance

1. Write the KVL equation for the super-mesh: $-40 + 4 \times 10^3 i_o + R \times 10^3 \times (3i_o) + 20 = 0$.

2. Simplify to get $(4 + 3R) \times 10^3 i_o = 20$.

3. Solve for $i_o$: $i_o = \frac{20}{4 + 3R}$ mA.

Try solving on your own before revealing the answer!

Final Answer: $i_o = \frac{20}{4 + 3R}$ mA

Substitute each value of R to find the current $i_o$ for each case.

Q4. Use KVL and KCL to determine the current $i_x$ in the circuit.

Background

Topic: Node and Loop Analysis, Dependent Sources

This question tests your ability to use Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to solve for a current in a circuit with dependent sources.

Key Terms and Formulas

• KCL: The sum of currents entering a node equals the sum leaving.

• KVL: The sum of voltages around a loop is zero.

• Dependent source: Source value depends on a circuit variable.

Step-by-Step Guidance

1. Assign currents: $i_1$ in the 3 Ω resistor, $i_2$ in the 2 Ω resistor, $i_x$ as shown.

2. Apply KCL: The current in the 6 Ω resistor is $6i_x$.

3. Write KVL for the left loop: $3i_1 = 3i_x + 6(6i_x) = 39i_x$ so $i_1 = 13i_x$.

4. Use KCL to find $i_2$: $i_2 = i_x + 13i_x = 14i_x$.

5. Write KVL for the big loop: $10 = 3i_1 + 2i_2 = 39i_x + 28i_x = 67i_x$.

Try solving on your own before revealing the answer!

Final Answer: $i_x = \frac{10}{67}$ A

Calculate $i_x$ using the final equation from the big loop.

Q5. Determine the equivalent resistor seen from terminals (a, b) for R = {4, 5.16, 8.31, 12} Ω.

Background

Topic: Series-Parallel Resistance, Equivalent Resistance

This question tests your ability to combine resistors in series and parallel to find the equivalent resistance between two terminals.

Key Terms and Formulas

• Parallel resistance: $R_p = \frac{R_1 R_2}{R_1 + R_2}$

• Series resistance: $R_s = R_1 + R_2$

Step-by-Step Guidance

1. Combine 60 Ω and 40 Ω in parallel: $R_{60||40} = \frac{60 \times 40}{60 + 40}$.

2. Combine 20 Ω and 30 Ω in parallel: $R_{20||30} = \frac{20 \times 30}{20 + 30}$.

3. Add R to $R_{20||30}$: $R_{sum} = R_{20||30} + R$.

4. Combine $R_{60||40}$ and $R_{sum}$ in parallel: $R_{ab} = \frac{R_{60||40} \times R_{sum}}{R_{60||40} + R_{sum}}$.

5. Simplify to $R_{ab} = \frac{24(12 + R)}{36 + R}$.

Try solving on your own before revealing the answer!

Final Answer: $R_{ab} = \frac{24(12 + R)}{36 + R}$ Ω

Substitute each value of R to find the equivalent resistance for each case.