KVL and KCL | Network Analysis

Tutorial 2 — Kirchhoff's Laws (KVL & KCL)

Worked problems applying Kirchhoff's Current Law (sum of currents at a node is zero) and Kirchhoff's Voltage Law (sum of voltages around a loop is zero), including circuits with dependent sources. Each problem is followed by a complete solution with the final result highlighted.

Network Analysis · Electric Circuit Analysis · 13 solved problems

Demonstrative VideoWalkthrough

Problem 1KCL & Dependent Source

Find \(v_0\) and \(i_0\) in the given circuit.

Solution

KCL at node \(A\): the 9 A source splits, so

\[ 9 = i_0 + i_1 \;\Rightarrow\; i_1 = 9 - i_0 \]

KCL at node \(B\), with \(i_2 = v_0/8\):

\[ i_1 = \frac{i_0}{4} + i_2 = \frac{i_0}{4} + \frac{v_0}{8} \]

Substituting the first relation into the second:

\[ \begin{aligned} 9 - i_0 &= \frac{i_0}{4} + \frac{v_0}{8}\\[2pt] 9 - \frac{v_0}{8} &= \frac{5 i_0}{4} \end{aligned} \]

The two nodes are at the same potential, so \(V_A = V_B = v_0 = 2 i_0\). Using this:

\[ \begin{aligned} 9 - \frac{v_0}{8} &= \frac{5 i_0}{4}\\[2pt] 9 - \frac{i_0}{4} &= \frac{5 i_0}{4} \;\Rightarrow\; 9 = \frac{6 i_0}{4}\\[2pt] i_0 &= 6\ \text{A},\qquad v_0 = 2 i_0 = 12\ \text{V} \end{aligned} \]

Answer\(i_0 = 6\ \text{A},\quad v_0 = 12\ \text{V}\)

Problem 2KVL + KCL

Obtain the current \(I\) in the network.

Solution

Applying KVL around Loop 2 (with a dependent term \(3V_R\)):

\[ 3V_R - 5I - 4 - V_R = 0 \;\Rightarrow\; 2V_R = 5I + 4 \]

KCL at node \(A\) gives \(I_1 = I - 2\), and the control voltage is

\[ V_R = 2 I_1 = 2I - 4 \]

Substituting into the loop equation:

\[ 2(2I - 4) = 5I + 4 \;\Rightarrow\; 4I - 8 = 5I + 4 \;\Rightarrow\; I = -12\ \text{A} \]

Answer\(I = -12\ \text{A}\)

Problem 3KVL & Dependent Source

Obtain the current \(i\) in the network.

Solution

KVL in the first loop fixes \(I_1\) and the node-\(a\) voltage:

\[ 5 = 2 I_1 \;\Rightarrow\; I_1 = 2.5\ \text{A},\qquad V_a = I_1 = 2.5\ \text{V} \]

KVL in the second loop relates the controlled source to \(i\):

\[ 4 V_{ab} = 4 i \;\Rightarrow\; i = V_{ab},\qquad V_b = i \]

Since \(V_{ab} = V_a - V_b\):

\[ V_{ab} = 2.5 - i = i \;\Rightarrow\; 2.5 = 2i \;\Rightarrow\; i = 1.25\ \text{A} \]

Answer\(i = 1.25\ \text{A}\)

Problem 4KVL — Branch Voltages

Determine the following in the circuit:

\(V_{R2}\)
\(V_2\), given \(V_{R1} = 1\)

Solution

aKVL in the first loop:

\[ -8 + 12 - V_{R2} = 0 \;\Rightarrow\; V_{R2} = 4\ \text{V} \]

Applying KVL through the next loop to find the helper voltage \(v_x\):

\[ V_{R2} - 7 + 9 - v_x = 0 \;\Rightarrow\; v_x = 6\ \text{V} \]

bKVL in the outer loop, with \(V_{R1} = 1\):

\[ v_x + V_2 + 3 - V_{R1} = 0 \;\Rightarrow\; V_2 = -v_x - 3 + V_{R1} = -6 - 3 + 1 = -8\ \text{V} \]

Answer\(V_{R2} = 4\ \text{V},\quad V_2 = -8\ \text{V}\)

Problem 5KCL at a Node

The voltage source in the given circuit drives 1 A out of its positive terminal into resistor \(R_1\). Calculate the current labelled \(i_2\).

Solution

Applying KCL at the node where the 1 A, the 3 A and the 7 A branches meet:

\[ 1 = i_2 - 3 + 7 \;\Rightarrow\; i_2 = -3\ \text{A} \]

Answer\(i_2 = -3\ \text{A}\)

Problem 6KCL — Node Voltage

Determine the value of the voltage \(v\).

Solution

Applying KCL at the node (currents in = currents out), with the two 5 Ω branches each carrying \(v/5\):

\[ \begin{aligned} 1 + 2 &= \frac{v}{5} + 5 + \frac{v}{5}\\[2pt] -2 &= \frac{2v}{5} \;\Rightarrow\; v = -5\ \text{V} \end{aligned} \]

Answer\(v = -5\ \text{V}\)

Problem 7KCL & Dependent Source

Find the current \(i_x\) in the circuit shown.

Solution

Applying KCL at the three nodes (current-controlled source \(3i_1\), resistors in kΩ):

\[ \begin{aligned} 4\,\text{mA} + 3 i_1 + \frac{V}{5\,\text{k}\Omega} + \frac{V}{20\,\text{k}\Omega} &= 0\\[2pt] i_1 &= 3 i_1 + \frac{V}{20\,\text{k}\Omega}\\[2pt] i_x &= 3 i_1 + \frac{V}{20\,\text{k}\Omega} \end{aligned} \]

Solving the first two equations for \(V\) and \(i_1\) and substituting into the third:

\[ i_x = 0.57\ \text{mA} \]

Unit check: because every source is in milliamps and every resistance in kilohms, the branch current is necessarily on the milliamp scale — \(i_x = 0.57\ \text{mA}\), not 0.57 A.

Answer\(i_x = 0.57\ \text{mA}\)

Problem 8KVL — Branch Voltages

For the given circuit, use KVL to find the branch voltages \(V_1\) through \(V_4\).

Solution

Applying KVL around the four loops:

\[ \begin{aligned} 3 - V_1 + V_3 &= 0\\ V_1 + V_2 + 2 &= 0\\ -4 - V_3 - V_4 &= 0\\ V_4 - 2 - 5 &= 0 \end{aligned} \]

Solving from the last equation upward:

\[ V_4 = 7\ \text{V},\quad V_3 = -4 - V_4 = -11\ \text{V},\quad V_1 = 3 + V_3 = -8\ \text{V},\quad V_2 = -V_1 - 2 = 6\ \text{V} \]

Answer\(V_1 = -8\ \text{V},\ V_2 = 6\ \text{V},\ V_3 = -11\ \text{V},\ V_4 = 7\ \text{V}\)

Problem 9KCL + Node Voltage

Determine the value of \(v_x\) labelled in the circuit.

Solution

Voltage at the first node (the 2.3 V source minus the drop produced by the 500 mA current in the 7.3 Ω resistor):

\[ V_1 = 2.3 - (0.5)(7.3) = -1.35\ \text{V} \]

KCL at the first node gives the branch current \(I_2\):

\[ I_2 = 0.5 + 1.35 = 1.85\ \text{A} \]

Finally, the drop across the 2 Ω resistor gives \(v_x\):

\[ v_x = -1.35 - (1.85)(2) = -5.05\ \text{V} \]

Answer\(v_x = -5.05\ \text{V}\)

Problem 10KCL — Node Equation

Find the current \(I_x\) in the circuit given.

Solution

Applying KCL at the node (both resistors are 100 Ω, source 10 mA):

\[ \frac{V - 1}{100} + \frac{V}{100} = 10\ \text{mA} \;\Rightarrow\; 2V - 1 = 1 \;\Rightarrow\; V = 1\ \text{V} \]

The required branch current:

\[ I_x = \frac{V}{100} = \frac{1}{100} = 10\ \text{mA} \]

Answer\(I_x = 10\ \text{mA}\)

Additional Practice Problems

Three further problems with self-contained schematics: a single-loop KVL with opposing sources, a KCL current-division node, and a two-loop circuit solved by the branch-current method (KVL + KCL together).

Problem 11Single-Loop KVL

In the single loop below, a 20 V source drives a clockwise current \(I\) through a 4 Ω and a 6 Ω resistor, opposed by a 5 V source. Find (a) the loop current \(I\), (b) the power delivered by the 20 V source, and (c) the voltage across the 6 Ω resistor.

Solution

aKVL clockwise (rise of 20 V, drops across the resistors, and the opposing 5 V source):

\[ 20 - 4I - 5 - 6I = 0 \;\Rightarrow\; 15 = 10 I \;\Rightarrow\; I = 1.5\ \text{A} \]

bPower delivered by the 20 V source:

\[ P = V I = 20 \times 1.5 = 30\ \text{W} \]

cVoltage across the 6 Ω resistor:

\[ V_{6\Omega} = 6 I = 6 \times 1.5 = 9\ \text{V} \]

Check: the drops \(4I + 5 + 6I = 6 + 5 + 9 = 20\,\text{V}\) balance the source — KVL is satisfied.

Answer\(I = 1.5\ \text{A},\ P = 30\ \text{W},\ V_{6\Omega} = 9\ \text{V}\)

Problem 12KCL & Current Division

A 6 A current source feeds node \(A\), which connects to ground through a 3 Ω and a 6 Ω resistor in parallel. Find the node voltage \(V_A\) and the two branch currents, and verify KCL.

Solution

The parallel resistance sets the node voltage. With \(R_p = (3\times 6)/(3+6) = 2\,\Omega\) and all 6 A flowing to ground:

\[ V_A = I \cdot R_p = 6 \times 2 = 12\ \text{V} \]

Branch currents by Ohm's law:

\[ I_{3\Omega} = \frac{12}{3} = 4\ \text{A},\qquad I_{6\Omega} = \frac{12}{6} = 2\ \text{A} \]

KCL check at node \(A\):

\[ I_{3\Omega} + I_{6\Omega} = 4 + 2 = 6\ \text{A} = I_{\text{source}} \checkmark \]

Answer\(V_A = 12\ \text{V},\ I_{3\Omega} = 4\ \text{A},\ I_{6\Omega} = 2\ \text{A}\)

Problem 13Branch-Current Method (KVL + KCL)

In the two-loop circuit, a 12 V source (with 3 Ω) and a 10 V source (with 4 Ω) share a common 2 Ω branch. Using branch currents \(I_1, I_2, I_3\), find all three currents.

Solution

KCL at the top node \(P\) (both source branches feed the middle branch):

\[ I_3 = I_1 + I_2 \]

KVL around the left loop (12 V source, 3 Ω, shared 2 Ω) and the right loop (10 V source, 4 Ω, shared 2 Ω):

\[ \begin{aligned} 12 &= 3 I_1 + 2 I_3 = 5 I_1 + 2 I_2\\ 10 &= 4 I_2 + 2 I_3 = 2 I_1 + 6 I_2 \end{aligned} \]

Solving the pair (from the second, \(I_1 = 5 - 3 I_2\); substitute into the first):

\[ 5(5 - 3 I_2) + 2 I_2 = 12 \;\Rightarrow\; -13 I_2 = -13 \;\Rightarrow\; I_2 = 1\ \text{A} \]

Back-substituting:

\[ I_1 = 5 - 3(1) = 2\ \text{A},\qquad I_3 = I_1 + I_2 = 3\ \text{A} \]

Check: left loop \(3(2) + 2(3) = 12\,\text{V}\); right loop \(4(1) + 2(3) = 10\,\text{V}\) — both KVL equations hold.

Answer\(I_1 = 2\ \text{A},\ I_2 = 1\ \text{A},\ I_3 = 3\ \text{A}\)