Chapter 3 Problems
Practice problems for Kirchhoff's Laws and circuit topology. Problems range from conceptual to computational, organized by topic.
KCL Problems
Problem 3.1
Five wires meet at a node. Currents in four of the wires are: 3 A (in), 7 A (out), 2 A (in), and 1 A (in). Find the current in the fifth wire and state its direction.
Show Solution
By KCL: \(\sum I_{in} = \sum I_{out}\). In: \(3 + 2 + 1 = 6\) A. Out: 7 A. Since more current leaves than enters, the fifth wire must carry \(7 - 6 = 1\) A into the node.
Problem 3.2
At a node, \(I_1 = 10\) mA enters and splits into three branches. If \(I_2 = 3\) mA and \(I_3 = 5\) mA, find \(I_4\).
Show Solution
\(I_4 = I_1 - I_2 - I_3 = 10 - 3 - 5 = 2\) mA leaving the node.
KVL Problems
Problem 3.3
A single loop contains a 24 V battery and four equal resistors. What is the voltage across each resistor?
Show Solution
By KVL, the total voltage drop equals the source: \(4V_R = 24\) V, so \(V_R = 6\) V across each resistor.
Problem 3.4
A loop has two voltage sources: \(V_1 = 10\) V and \(V_2 = 4\) V (opposing), and two resistors \(R_1 = 3\;\Omega\) and \(R_2 = 3\;\Omega\). Find the loop current and verify KVL.
Show Solution
Net voltage: \(10 - 4 = 6\) V. Total resistance: \(3 + 3 = 6\;\Omega\). Current: \(I = 6/6 = 1\) A.
KVL check: \(+10 - (1)(3) - (1)(3) - 4 = 10 - 3 - 3 - 4 = 0\) ✓
Topology Problems
Problem 3.5
A circuit has 3 nodes and 4 branches. (a) How many independent KCL equations can you write? (b) How many independent meshes are there?
Show Solution
(a) KCL equations: \(n - 1 = 3 - 1 = 2\).
(b) Using \(b = n - 1 + m\): \(4 = 2 + m\), so \(m = 2\) meshes.
Node Voltage & Mesh Current Problems
Problem 3.6
A 20 V source connects to node A through \(R_1 = 5\;\text{k}\Omega\). Node A connects to ground through \(R_2 = 10\;\text{k}\Omega\). Use the node voltage method to find \(V_A\).
Show Solution
KCL at node A: \(\frac{V_A - 20}{5\text{k}} + \frac{V_A}{10\text{k}} = 0\). Multiply by 10k: \(2(V_A - 20) + V_A = 0\), so \(3V_A = 40\), giving \(V_A = 13.33\) V. This is simply a voltage divider: \(V_A = 20 \times \frac{10}{5+10} = 13.33\) V.
Problem 3.7
Two meshes share a \(4\;\Omega\) resistor. Mesh 1 contains a 10 V source and a \(2\;\Omega\) resistor. Mesh 2 contains a \(6\;\Omega\) resistor. Find both mesh currents.
Show Solution
Mesh 1: \(-10 + 2I_1 + 4(I_1 - I_2) = 0 \;\Rightarrow\; 6I_1 - 4I_2 = 10\)
Mesh 2: \(4(I_2 - I_1) + 6I_2 = 0 \;\Rightarrow\; -4I_1 + 10I_2 = 0\)
From equation 2: \(I_1 = 2.5 I_2\). Substituting: \(6(2.5 I_2) - 4I_2 = 10\), so \(11I_2 = 10\), giving \(I_2 = 0.909\) A and \(I_1 = 2.273\) A.
Delta-Wye Problems
Problem 3.8
Convert a delta configuration with \(R_a = 12\;\Omega\), \(R_b = 6\;\Omega\), \(R_c = 18\;\Omega\) to its equivalent wye.
Show Solution
\(\sum R_\Delta = 12 + 6 + 18 = 36\;\Omega\)
\(R_1 = \frac{R_a R_b}{\sum} = \frac{12 \times 6}{36} = 2\;\Omega\)
\(R_2 = \frac{R_b R_c}{\sum} = \frac{6 \times 18}{36} = 3\;\Omega\)
\(R_3 = \frac{R_a R_c}{\sum} = \frac{12 \times 18}{36} = 6\;\Omega\)
Conceptual Problems
Problem 3.9
Explain why superposition cannot be used to calculate power directly, even though it works for voltage and current.
Show Solution
Superposition relies on linearity. Power is proportional to \(I^2\) or \(V^2\), which are nonlinear functions. The square of a sum is not equal to the sum of squares: \((I_1 + I_2)^2 \neq I_1^2 + I_2^2\). You must first find the total current or voltage using superposition, then calculate power from the combined result.
Problem 3.10
A circuit has 6 nodes and you want to minimize the number of equations. Would you use the node voltage method or the mesh current method? Justify your choice.
Show Solution
The node voltage method requires \(n - 1 = 5\) equations. Without knowing the mesh count, you cannot be certain. However, if the circuit is planar and has fewer meshes than 5, the mesh method would be better. If the circuit has many parallel branches (few meshes), use mesh analysis. If it has many series elements (few nodes relative to meshes), use node voltage analysis. The general rule: choose whichever produces fewer equations.