Training Electrical Engineering DC Circuit Analysis — Kirchhoff's Laws
1 / 5

DC Circuit Analysis — Kirchhoff's Laws

24 min Electrical Engineering

DC Circuit Analysis — Kirchhoff's Laws

Circuit analysis is the backbone of electrical engineering. Kirchhoff's laws let us solve for unknown voltages and currents in any circuit.

Kirchhoff's Current Law (KCL)

The sum of currents entering a node equals the sum leaving:

$$\sum I_{\text{in}} = \sum I_{\text{out}}$$

Kirchhoff's Voltage Law (KVL)

The sum of voltage drops around any closed loop is zero:

$$\sum V = 0$$

Ohm's Law

$$V = IR$$

Voltage equals current times resistance. Power: $P = IV = I^2R = V^2/R$.

Example 1

A circuit has a 12 V battery and two resistors in series: $R_1 = 4\,\Omega$, $R_2 = 8\,\Omega$. Find the current and voltage across each resistor.

$$I = \frac{V}{R_1 + R_2} = \frac{12}{4 + 8} = 1 \text{ A}$$

$V_1 = IR_1 = 4$ V, $V_2 = IR_2 = 8$ V. Check: $4 + 8 = 12$ V ✓

Example 2

Two resistors $R_1 = 6\,\Omega$ and $R_2 = 3\,\Omega$ are in parallel across a 9 V source. Find the total current.

$$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{6 \times 3}{6 + 3} = 2\,\Omega$$

$$I_{\text{total}} = \frac{9}{2} = 4.5 \text{ A}$$

$I_1 = 9/6 = 1.5$ A, $I_2 = 9/3 = 3$ A. KCL check: $1.5 + 3 = 4.5$ ✓

Example 3

A circuit has two loops. Loop 1: $10 - 2I_1 - 3(I_1 - I_2) = 0$. Loop 2: $-3(I_2 - I_1) - 4I_2 + 5 = 0$. Find $I_1$ and $I_2$.

Simplify Loop 1: $5I_1 - 3I_2 = 10$

Simplify Loop 2: $-3I_1 + 7I_2 = 5$

From Cramer's rule or substitution: $I_1 = 3.27$ A, $I_2 = 2.12$ A

Practice Problems

1. Three resistors in series: 10 Ω, 20 Ω, 30 Ω with a 24 V source. Find the current and each voltage drop.
2. Three resistors in parallel: 12 Ω, 6 Ω, 4 Ω. Find $R_{\text{eq}}$.
3. A 100 W light bulb operates at 120 V. Find its resistance and current.
4. Apply KVL: $V_s - V_{R1} - V_{R2} - V_{R3} = 0$. If $V_s = 15$ V, $V_{R1} = 3$ V, $V_{R3} = 7$ V, find $V_{R2}$.
5. A node has 3 A and 5 A entering, and 4 A leaving on one branch. Find the current in the other branch.
6. Two 10 Ω resistors in parallel, then in series with a 5 Ω resistor. Total voltage = 20 V. Find the current.
7. A Wheatstone bridge has $R_1 = 100$, $R_2 = 200$, $R_3 = 150$, $R_4 = ?$ for balance. Find $R_4$.
8. Find the power dissipated in a 50 Ω resistor carrying 2 A.
9. A circuit has two voltage sources: 12 V and 8 V opposing, series resistance 10 Ω. Find the current.
10. Use superposition: two sources, $V_1 = 10$ V with $R_1 = 5$ Ω, $V_2 = 6$ V with $R_2 = 3$ Ω, shared $R_L = 15$ Ω.
11. A voltage divider: $V_{\text{out}} = V_{\text{in}} \cdot R_2/(R_1 + R_2)$. Design for $V_{\text{out}} = 3.3$ V from $V_{\text{in}} = 5$ V using $R_2 = 10$ kΩ.
12. Find the Thévenin equivalent voltage and resistance seen by $R_L$ in a circuit with $V_s = 20$ V, $R_1 = 4$ Ω, $R_2 = 6$ Ω.
Show Answer Key

1. $I = 24/60 = 0.4$ A; $V_1=4$ V, $V_2=8$ V, $V_3=12$ V

2. $1/R_{eq} = 1/12 + 1/6 + 1/4 = 1/2$; $R_{eq} = 2$ Ω

3. $R = V^2/P = 14{,}400/100 = 144$ Ω; $I = 120/144 = 0.833$ A

4. $V_{R2} = 15 - 3 - 7 = 5$ V

5. $I_{\text{out}} = 3 + 5 - 4 = 4$ A

6. $R_{\text{parallel}} = 5$ Ω; $R_{\text{total}} = 10$ Ω; $I = 2$ A

7. Balance: $R_1/R_2 = R_3/R_4$; $R_4 = 200 \times 150/100 = 300$ Ω

8. $P = I^2R = 4 \times 50 = 200$ W

9. Net voltage $= 12 - 8 = 4$ V; $I = 4/10 = 0.4$ A

10. By superposition: $I_{L1} = 10/(5+15) = 0.5$ A, $I_{L2} = 6/(3+15) = 0.333$ A; total depends on polarities.

11. $R_1 = R_2(V_{\text{in}}/V_{\text{out}} - 1) = 10{,}000 \times (5/3.3 - 1) \approx 5.15$ kΩ

12. $V_{th} = 20 \times 6/(4+6) = 12$ V; $R_{th} = 4 \times 6/(4+6) = 2.4$ Ω

AC Circuits — Phasors & Impedance