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Chapter 2 Practice Problems — Ohm's Law and Basic Configurations

These problems are meant to build your problem-solving intuition. A hint is provided for each — try the problem on your own first before reading it.

Problem 1 — Ohm's Law and Power

A 12 V source is connected to a single resistor and the current is measured at 40 mA.

(a) Find the resistance using Ohm's Law.

(b) Calculate the power dissipated by the resistor using all three power formulas (\(P = VI\), \(P = I^2R\), \(P = V^2/R\)) and verify they agree.

(c) Is a standard 1/4 W (250 mW) resistor adequate for this application? What rating would you choose?

Hint

(a) Rearrange Ohm's Law to \(R = V/I\). Convert mA to A: 40 mA = 0.040 A.

(b) All three formulas must yield the same power. If they don't, check your arithmetic. Express the final answer in milliwatts.

(c) Compare the computed power to 250 mW. A safe design practice is to derate to 50–70% of the rated maximum, meaning the actual dissipation should be well below the rated value. Choose the next standard rating above your calculated power.

Problem 2 — Series Resistors

Three resistors — \(R_1 = 1.0\) kΩ, \(R_2 = 2.2\) kΩ, and \(R_3 = 3.3\) kΩ — are connected in series across an 18 V supply.

(a) Find the total equivalent resistance.

(b) Find the current through the series string.

(c) Find the voltage drop across each resistor. Verify that the three voltage drops sum to 18 V (KVL check).

Hint

(a) Series resistors add directly: \(R_{total} = R_1 + R_2 + R_3\). Keep all values in kΩ and the answer will also be in kΩ.

(b) With the total resistance found, apply \(I = V/R_{total}\). The same current flows through every element in series.

(c) Apply Ohm's Law to each resistor: \(V_k = I \cdot R_k\). A larger resistor always has a larger voltage drop when the current is common.

Problem 3 — Parallel Resistors and Current Divider

Two resistors, \(R_1 = 6\) kΩ and \(R_2 = 12\) kΩ, are connected in parallel across a 9 V supply.

(a) Find the equivalent parallel resistance using the product-over-sum formula.

(b) Find the total current supplied by the source.

(c) Use the current divider formula to find the current through each resistor. Verify the currents add to the total found in (b).

Hint

(a) For two resistors in parallel: \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2}\). The result is always smaller than the smaller of the two resistors.

(b) \(I_{total} = V / R_{eq}\). Since both resistors share the same 9 V, you can also confirm by computing \(V/R_1\) and \(V/R_2\) separately and adding.

(c) Current divider: \(I_1 = I_{total} \cdot \frac{R_2}{R_1 + R_2}\). Note that more current flows through the smaller resistor — this is the path of least resistance.

Problem 4 — Voltage Divider with a Load

A voltage divider is built from \(R_1 = 10\) kΩ (top) and \(R_2 = 10\) kΩ (bottom), supplied from \(V_{in} = 10\) V.

(a) Find the unloaded output voltage \(V_{out}\) (no load connected).

(b) A load \(R_L = 10\) kΩ is now connected across \(R_2\). Find the effective parallel resistance of \(R_2 \| R_L\), then calculate the new \(V_{out}\).

(c) What is the percentage drop in output voltage due to the load? Why is this load poorly matched to this divider?

Hint

(a) Apply the voltage divider formula: \(V_{out} = V_{in} \cdot \frac{R_2}{R_1 + R_2}\). With equal resistors the answer is exactly half the supply.

(b) Compute \(R_2 \| R_L = \frac{R_2 \cdot R_L}{R_2 + R_L}\), then substitute this effective resistance as the new "bottom" resistor in the divider formula.

(c) Percentage drop = \(\frac{V_{unloaded} - V_{loaded}}{V_{unloaded}} \times 100\%\). For a divider to work well under load, the load resistance should be at least 10× the bottom resistor value.