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Chapter 2 Quiz — Ohm's Law and Basic Circuit Configurations

Test your understanding of Ohm's Law, series and parallel resistor combinations, voltage and current dividers, and power calculations. Each question includes a detailed explanation.

Question 1

Three resistors — 2 kΩ, 3 kΩ, and 5 kΩ — are connected in series across a 20 V source. What is the current flowing through the circuit?

  1. 10 mA
  2. 4 mA
  3. 2 mA
  4. 20 mA
Show Answer

Correct Answer: C

Series resistors add: \(R_{total} = 2 + 3 + 5 = 10 \text{ k}\Omega\). Then by Ohm's Law: \(I = V/R_{total} = 20 \text{ V} / 10{,}000 \text{ }\Omega = 0.002 \text{ A} = 2 \text{ mA}\). A key property of series circuits: the same current flows through every element.

Concept Tested: Series Resistance \(R_{eq} = R_1 + R_2 + \cdots\)

Question 2

Two resistors, 6 kΩ and 3 kΩ, are connected in parallel. What is their equivalent resistance?

  1. 9 kΩ
  2. 4.5 kΩ
  3. 2 kΩ
  4. 1 kΩ
Show Answer

Correct Answer: C

Using the product-over-sum formula for two parallel resistors: \(R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \text{ k}\Omega\). Notice the result is less than either individual resistor — a key property of parallel combinations. Adding parallel paths always reduces overall resistance.

Concept Tested: Parallel Resistance \(R_{eq} = R_1 R_2 / (R_1 + R_2)\)

Question 3

In a voltage divider, \(V_{in} = 15 \text{ V}\), \(R_1 = 10 \text{ k}\Omega\) (top resistor), and \(R_2 = 5 \text{ k}\Omega\) (bottom resistor connected to ground). What is \(V_{out}\) (the voltage at the junction of R1 and R2)?

  1. 10 V
  2. 7.5 V
  3. 5 V
  4. 3 V
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Correct Answer: C

Using the voltage divider formula: \(V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2} = 15 \times \frac{5}{10 + 5} = 15 \times \frac{1}{3} = 5 \text{ V}\). The output voltage is the fraction of input determined by the ratio of the bottom resistor to the total. When \(R_2\) is one-third of the total, the output is one-third of the input.

Concept Tested: Voltage Divider Formula

Question 4

A 100 mA total current enters a parallel combination of \(R_1 = 4 \text{ k}\Omega\) and \(R_2 = 1 \text{ k}\Omega\). How much current flows through \(R_1\)?

  1. 80 mA
  2. 50 mA
  3. 25 mA
  4. 20 mA
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Correct Answer: D

Using the current divider formula: \(I_1 = I_{total} \cdot \frac{R_2}{R_1 + R_2} = 100 \times \frac{1}{4 + 1} = 100 \times \frac{1}{5} = 20 \text{ mA}\). Notice that the current through \(R_1\) depends on \(R_2\) in the numerator — more current flows through the lower resistance branch. \(R_1\) has 4× the resistance of \(R_2\), so it carries only 1/5 of the total current while \(R_2\) carries 4/5.

Concept Tested: Current Divider Formula

Question 5

A resistor has 6 V across it and 3 mA through it. How much power does it dissipate?

  1. 2 mW
  2. 9 mW
  3. 18 mW
  4. 2 W
Show Answer

Correct Answer: C

Using \(P = VI = 6 \text{ V} \times 3 \times 10^{-3} \text{ A} = 18 \times 10^{-3} \text{ W} = 18 \text{ mW}\). This is the most direct form when both V and I are known. You can verify the resistance is \(R = V/I = 6/0.003 = 2 \text{ k}\Omega\), and check \(P = I^2 R = (0.003)^2 \times 2000 = 18 \text{ mW}\). All three power formulas give the same answer.

Concept Tested: Power Calculation \(P = VI\)

Question 6

A circuit has a 24 V source, with R1 = 2 kΩ in series with a parallel combination of R2 = 6 kΩ and R3 = 3 kΩ. What is the total current drawn from the source?

  1. 4 mA
  2. 6 mA
  3. 8 mA
  4. 12 mA
Show Answer

Correct Answer: B

Step 1: Find \(R_{23}\) (parallel): \(R_{23} = (6 \times 3)/(6 + 3) = 18/9 = 2 \text{ k}\Omega\). Step 2: Total resistance: \(R_{total} = R_1 + R_{23} = 2 + 2 = 4 \text{ k}\Omega\). Step 3: Total current: \(I = 24/4000 = 6 \text{ mA}\). The voltage across the parallel combination is \(6 \text{ mA} \times 2 \text{ k}\Omega = 12 \text{ V}\), and voltages add: \(12 + 12 = 24\) V. ✓

Concept Tested: Series-Parallel Circuit Analysis

Question 7

A resistor is rated at 1/4 W (0.25 W). If 5 V is applied across it, what is the minimum resistance it can safely have?

  1. 25 Ω
  2. 100 Ω
  3. 1.25 kΩ
  4. 500 Ω
Show Answer

Correct Answer: B

Set power equal to the rating and solve for R: \(P = V^2/R\), so \(R = V^2/P = (5)^2/0.25 = 25/0.25 = 100 \text{ }\Omega\). Any resistance lower than 100 Ω would dissipate more than 0.25 W and risk damaging the resistor. For safety, you would derate to 50–70% of the rating and choose an even higher value resistor.

Concept Tested: Component Power Rating

Question 8

A voltage divider uses R1 = 10 kΩ and R2 = 10 kΩ across a 5 V source. A 10 kΩ load resistor is then connected in parallel with R2. What is the output voltage across the load?

  1. 2.5 V
  2. 3.33 V
  3. 1.67 V
  4. 5 V
Show Answer

Correct Answer: C

The 10 kΩ load parallels R2 (10 kΩ): \(R_{2,eff} = 10k \| 10k = 5 \text{ k}\Omega\). Now the divider is R1 = 10 kΩ, \(R_{2,eff}\) = 5 kΩ: \(V_{out} = 5 \times \frac{5}{10+5} = 5 \times \frac{1}{3} = 1.67 \text{ V}\). Without the load it was 2.5 V — the load dragged it down to 1.67 V, a 33% reduction. This is the loading effect, and it is why high-impedance inputs matter.

Concept Tested: Voltage Divider Loading Effect

Answers Summary

Question Answer Concept Tested
1 C Series Resistance
2 C Parallel Resistance
3 C Voltage Divider Formula
4 D Current Divider Formula
5 C Power Calculation
6 B Series-Parallel Circuit Analysis
7 B Component Power Rating
8 C Voltage Divider Loading Effect