Chapter 2 Practice Problems
Practice Problems
Problem 1 — Ohm's Law and Power
A 9V battery is connected to a 470Ω resistor.
(a) Find the current through the resistor.
(b) Find the power dissipated by the resistor.
(c) Is a standard 1/4W (0.25W) resistor adequate?
(d) What value of resistance would dissipate exactly 1W at 9V?
Solution
(a) [I = \frac{V}{R} = \frac{9\text{ V}}{470\text{ Ω}} = 19.1\text{ mA}]
(b) [P = I^2 R = (0.0191)^2 \times 470 = 0.000365 \times 470 = 171.6\text{ mW}]
Or: \(P = V^2/R = 81/470 = 172\text{ mW}\)
(c) 172 mW > 250 mW? No — 172 mW is less than 250 mW, so a 1/4W resistor is adequate. However, it operates at 69% of rating, which is marginal. A 1/2W resistor would be better practice (34% utilization).
(d) [R = \frac{V^2}{P} = \frac{81}{1} = 81\text{ Ω}]
Problem 2 — Series Circuit Analysis
A 24V source is connected to three series resistors: R1 = 2kΩ, R2 = 4kΩ, R3 = 6kΩ.
(a) Find the equivalent resistance.
(b) Find the current through the circuit.
(c) Find the voltage across each resistor.
(d) Verify Kirchhoff's Voltage Law.
Solution
(a) [R_{eq} = R_1 + R_2 + R_3 = 2\text{k} + 4\text{k} + 6\text{k} = 12\text{kΩ}]
(b) [I = \frac{V}{R_{eq}} = \frac{24\text{ V}}{12{,}000\text{ Ω}} = 2\text{ mA}]
(c) [V_1 = IR_1 = 2\text{ mA} \times 2\text{kΩ} = 4\text{ V}] [V_2 = IR_2 = 2\text{ mA} \times 4\text{kΩ} = 8\text{ V}] [V_3 = IR_3 = 2\text{ mA} \times 6\text{kΩ} = 12\text{ V}]
(d) KVL check: \(V_1 + V_2 + V_3 = 4 + 8 + 12 = 24\text{ V} = V_s\) ✓
Problem 3 — Parallel Circuit Analysis
Three resistors R1 = 100Ω, R2 = 200Ω, and R3 = 400Ω are connected in parallel across a 20V source.
(a) Find the equivalent resistance.
(b) Find the current through each resistor.
(c) Find the total current from the source.
(d) Verify: the total current should equal \(V_s / R_{eq}\).
Solution
(a) [\frac{1}{R_{eq}} = \frac{1}{100} + \frac{1}{200} + \frac{1}{400} = \frac{4}{400} + \frac{2}{400} + \frac{1}{400} = \frac{7}{400}]
(b) All resistors share the 20V source voltage:
(c) [I_{total} = I_1 + I_2 + I_3 = 200 + 100 + 50 = 350\text{ mA}]
(d) [I_{total} = \frac{V_s}{R_{eq}} = \frac{20}{400/7} = \frac{20 \times 7}{400} = \frac{140}{400} = 0.35\text{ A} = 350\text{ mA}] ✓
Problem 4 — Voltage Divider Design
You need to produce a 3.3V reference from a 12V supply using a voltage divider. Assume the load draws negligible current.
(a) If R1 = 10kΩ, find the required value of R2.
(b) Calculate the current flowing through the divider.
(c) What power does each resistor dissipate?
(d) Are 1/4W resistors adequate?
Solution
(a) Using the voltage divider formula and solving for R2:
The nearest standard value is 3.9kΩ (E24 series), giving \(V_{out} = 12 \times 3.9/(10 + 3.9) = 3.37\text{ V}\), which is close enough for most applications.
(b) [I = \frac{V_{in}}{R_1 + R_2} = \frac{12\text{ V}}{10\text{k} + 3.79\text{k}} = \frac{12}{13.79\text{k}} = 0.87\text{ mA}]
(c) [P_1 = I^2 R_1 = (0.00087)^2 \times 10{,}000 = 7.57\text{ mW}] [P_2 = I^2 R_2 = (0.00087)^2 \times 3{,}790 = 2.87\text{ mW}]
(d) Both are well below 250 mW. Yes, 1/4W resistors are more than adequate.