Skip to content

Chapter 3 Practice Problems

Practice Problems

Problem 1 — KCL at Multiple Nodes

At a node in a circuit, five branch currents are present: - \(I_1 = 6\) A entering - \(I_2 = 2\) A entering - \(I_3 = 5\) A leaving - \(I_4 = ?\) leaving - \(I_5 = 1\) A entering

(a) Write the KCL equation for this node.

(b) Solve for \(I_4\).

(c) Verify the result by checking that the sum of all currents equals zero.

Solution

(a) KCL states: sum of currents entering = sum of currents leaving

\[\text{Entering: } I_1 + I_2 + I_5 = \text{Leaving: } I_3 + I_4\]

(b) Substituting known values:

[6 + 2 + 1 = 5 + I_4] [9 = 5 + I_4] [I_4 = 4 \text{ A (leaving the node)}]

(c) Verification (entering positive, leaving negative):

\[6 + 2 + 1 - 5 - 4 = 0 \quad \checkmark\]

Problem 2 — KVL in a Multi-Source Loop

A single loop circuit contains the following elements in series (traversed clockwise): - 18 V voltage source (+ terminal encountered first → rise) - \(R_1 = 3\) Ω - 6 V voltage source (− terminal encountered first → rise) - \(R_2 = 6\) Ω

(a) Write the KVL equation around the loop.

(b) Solve for the loop current \(I\).

(c) Find the voltage across \(R_2\).

Solution

(a) Traversing clockwise, applying KVL (rises positive, drops negative):

\[+18 - I(3) + 6 - I(6) = 0\]

Note: The 6 V source with − terminal first means entering the − side and exiting the + side → voltage rise of +6 V.

(b) Solving for \(I\):

[24 - 9I = 0] [I = \frac{24}{9} = 2.67 \text{ A}]

(c) Voltage across \(R_2\):

\[V_{R_2} = I \times R_2 = 2.67 \times 6 = 16 \text{ V}\]

Problem 3 — Node Voltage Method

A circuit has three nodes: ground (reference), Node 1, and Node 2. Component connections: - 12 V source between Node 1 and ground (Node 1 is positive terminal) - \(R_1 = 6\) Ω between Node 1 and Node 2 - \(R_2 = 4\) Ω between Node 2 and ground - \(R_3 = 12\) Ω between Node 2 and ground

(a) Identify the known and unknown node voltages.

(b) Write the KCL equation at Node 2.

(c) Solve for \(V_2\).

(d) Find the current through \(R_1\).

Solution

(a) Known: \(V_1 = 12\) V (set directly by the voltage source). Unknown: \(V_2\).

(b) KCL at Node 2 — sum of currents leaving Node 2 equals zero:

\[\frac{V_2 - V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_2}{R_3} = 0\]
\[\frac{V_2 - 12}{6} + \frac{V_2}{4} + \frac{V_2}{12} = 0\]

(c) Multiply through by 12 (LCD):

[2(V_2 - 12) + 3V_2 + V_2 = 0] [2V_2 - 24 + 3V_2 + V_2 = 0] [6V_2 = 24] [V_2 = 4 \text{ V}]

(d) Current through \(R_1\) (from Node 1 to Node 2):

\[I_{R_1} = \frac{V_1 - V_2}{R_1} = \frac{12 - 4}{6} = \frac{8}{6} = 1.33 \text{ A}\]

Problem 4 — Mesh Current Method

A circuit has two meshes: - Mesh 1: 10 V source, \(R_1 = 2\) Ω (shared with Mesh 2), \(R_2 = 3\) Ω - Mesh 2: \(R_1 = 2\) Ω (shared), \(R_3 = 4\) Ω

Let \(I_1\) be the clockwise mesh current in Mesh 1, and \(I_2\) be the clockwise mesh current in Mesh 2. There is no source in Mesh 2.

(a) Write the KVL mesh equation for Mesh 1.

(b) Write the KVL mesh equation for Mesh 2.

(c) Solve the system of equations for \(I_1\) and \(I_2\).

(d) Find the actual current through the shared resistor \(R_1\).

Solution

(a) Mesh 1 KVL (clockwise, drops negative):

[10 - I_1(R_2) - (I_1 - I_2)(R_1) = 0] [10 - 3I_1 - 2(I_1 - I_2) = 0] [10 - 5I_1 + 2I_2 = 0 \quad \text{...(1)}]

(b) Mesh 2 KVL (no source, drops only):

[-(I_2 - I_1)(R_1) - I_2(R_3) = 0] [-2(I_2 - I_1) - 4I_2 = 0] [2I_1 - 6I_2 = 0 \quad \text{...(2)}]

(c) From equation (2): \(I_1 = 3I_2\)

Substitute into equation (1):

[10 - 5(3I_2) + 2I_2 = 0] [10 - 13I_2 = 0] [I_2 = \frac{10}{13} \approx 0.769 \text{ A}] [I_1 = 3I_2 = \frac{30}{13} \approx 2.31 \text{ A}]

(d) The current through \(R_1\) is the difference of the two mesh currents (Mesh 1 flows down through \(R_1\), Mesh 2 flows up):

\[I_{R_1} = I_1 - I_2 = \frac{30}{13} - \frac{10}{13} = \frac{20}{13} \approx 1.54 \text{ A}\]

Problem 5 — Delta-to-Wye Conversion

Three resistors are connected in a delta configuration: \(R_{ab} = 30\) Ω, \(R_{bc} = 60\) Ω, \(R_{ca} = 90\) Ω.

(a) Convert to an equivalent wye configuration. Find \(R_a\), \(R_b\), and \(R_c\).

(b) Verify that the sum \(R_a + R_b + R_c\) is less than the smallest delta resistor (a quick sanity check for the conversion).

Solution

(a) Delta-to-wye conversion formulas. Let the sum of delta resistors be:

\[\Sigma R_\Delta = R_{ab} + R_{bc} + R_{ca} = 30 + 60 + 90 = 180 \text{ Ω}\]
\[R_a = \frac{R_{ab} \cdot R_{ca}}{\Sigma R_\Delta} = \frac{30 \times 90}{180} = \frac{2700}{180} = 15 \text{ Ω}\]
\[R_b = \frac{R_{ab} \cdot R_{bc}}{\Sigma R_\Delta} = \frac{30 \times 60}{180} = \frac{1800}{180} = 10 \text{ Ω}\]
\[R_c = \frac{R_{bc} \cdot R_{ca}}{\Sigma R_\Delta} = \frac{60 \times 90}{180} = \frac{5400}{180} = 30 \text{ Ω}\]

(b) Sanity check:

\[R_a + R_b + R_c = 15 + 10 + 30 = 55 \text{ Ω}\]

The smallest delta resistor is 30 Ω, and 55 Ω > 30 Ω, so this particular check doesn't apply here. However, each individual wye resistor (15, 10, 30 Ω) is smaller than at least two of the delta resistors — which is the correct expectation for a delta-to-wye conversion.