GTU 3130906 Electrical Circuit Analysis Summer 2026 Solved Paper & Solutions

Thevenin's theorem: any linear two terminal network can be replaced by a single voltage source $V_{th}$ in series with a resistance $R_{th}$. $V_{th}$ is the open circuit voltage at the terminals, and $R_{th}$ is the resistance seen from the terminals with all independent sources killed.

Norton's theorem: the same network can be replaced by a single current source $I_N$ in parallel with a resistance $R_N$. $I_N$ is the short circuit current at the terminals, and $R_N = R_{th}$.

The two are linked by $I_N = V_{th}/R_{th}$.

The short circuit admittance (y) parameters express the port currents in terms of the port voltages. Each one is found with one port short circuited.

• $y_{11} = I_1/V_1$ with $V_2 = 0$: input admittance.
• $y_{12} = I_1/V_2$ with $V_1 = 0$: reverse transfer admittance.
• $y_{21} = I_2/V_1$ with $V_2 = 0$: forward transfer admittance.
• $y_{22} = I_2/V_2$ with $V_1 = 0$: output admittance.

The y parameters are also called the short circuit admittance parameters, and they are the natural choice for networks joined in parallel. All four have the unit siemens.

Equivalent circuit: each equation is a KCL at one port. At port 1 the admittance $y_{11}$ is in parallel with a current source $y_{12}V_2$. At port 2 the admittance $y_{22}$ is in parallel with a current source $y_{21}V_1$.

For a reciprocal network $y_{12} = y_{21}$, and the circuit becomes the simple $\pi$ network with $-y_{12}$ in the series arm.

The 2 + j3 $\Omega$ branch joins the nodes $A$ (after the 5 $\Omega$ resistor) and $B$ (before the 4 $\Omega$ resistor). If the current in it is zero, there is no voltage drop, so $V_A = V_B$. Take both sources as shown, with the same polarity.

Check: $V_A = 21.2\angle 45^\circ$ V, and $V_2 = 35.36\angle 45^\circ$ V, so $V_2$ is in phase with $V_A$ and $0.6V_2 = 21.2\angle 45^\circ$ V, which equals $V_A$.

Statement: A load takes the maximum power from a source when its impedance is matched to the source impedance. For a DC circuit the condition is $R_L = R_{th}$.

For AC circuits, the source is $V_{th}$ with impedance $Z_{th} = R_{th} + jX_{th}$, and the load is $Z_L = R_L + jX_L$. The load current is

The real part of the load takes the power, and the reactance only swings energy. So the load reactance must cancel the source reactance, and the load resistance must then match the source resistance. The exact condition depends on which parts of the load can be changed:

1. Both $R_L$ and $X_L$ variable: $X_L = -X_{th}$ (to cancel the reactance) and $R_L = R_{th}$. So $Z_L = Z_{th}^*$, the complex conjugate. Then $P_{max} = \dfrac{V_{th}^2}{4R_{th}}$.
2. Only $R_L$ variable ($X_L$ fixed): $R_L = \sqrt{R_{th}^2 + (X_{th} + X_L)^2}$.
3. Only the magnitude $|Z_L|$ variable (power factor angle fixed): $|Z_L| = |Z_{th}|$.
4. A purely resistive load: $R_L = |Z_{th}| = \sqrt{R_{th}^2 + X_{th}^2}$.

Condition 1 gives the largest possible power of all.

Statement: In a linear bilateral network containing more than one independent source, the response (current or voltage) in any branch is the algebraic sum of the responses produced by each independent source acting alone, while all the other independent sources are replaced by their internal resistances.

Replacing the sources: an ideal voltage source becomes a short circuit and an ideal current source becomes an open circuit. A dependent source is not removed, because it depends on a circuit variable.

Steps:

1. Keep one source, remove the others, and solve the simpler circuit.
2. Repeat for each source.
3. Add the individual responses, with their directions.

Limits: it holds only for linear networks, so it can be used for current and voltage but not for power, as power is proportional to the square of the current.

Example: A 12 V source is in series with 2 $\Omega$, feeding a node $A$. A 4 $\Omega$ resistor joins $A$ to ground, and a 3 A current source feeds $A$. Find the current $I$ in the 4 $\Omega$ resistor.

• 12 V acting alone (current source open): $I' = \dfrac{12}{2 + 4} = 2$ A.
• 3 A acting alone (voltage source shorted): the current divides between 2 $\Omega$ and 4 $\Omega$, so $I'' = 3\times\dfrac{2}{2 + 4} = 1$ A.
• Total: $I = I' + I'' = 3$ A.

Check by nodal analysis: $\dfrac{V_A - 12}{2} + \dfrac{V_A}{4} = 3$ gives $V_A = 12$ V, and $I = 12/4 = 3$ A.

In the series-parallel connection of two two-port networks, the inputs are connected in series and the outputs are connected in parallel.

• The input current $I_1$ is the same in both networks, and the input voltages add.
• The output voltage $V_2$ is the same across both, and the output currents add.

So the h (hybrid) parameters add:

The h parameters are the natural choice here, as they have $I_1$ and $V_2$ as the independent variables.

Kirchhoff's Current Law (KCL): the algebraic sum of the currents at a node is zero, or the current entering a node equals the current leaving it. $\sum I = 0$. It follows from the conservation of charge.

Kirchhoff's Voltage Law (KVL): the algebraic sum of the voltages around any closed loop is zero, or the sum of the rises equals the sum of the drops. $\sum V = 0$. It follows from the conservation of energy.

Example of KCL: at a node, 5 A comes in, and 2 A and $I$ go out. Then $5 = 2 + I$, so $I = 3$ A.

Example of KVL: a 12 V battery is in series with 4 $\Omega$ and 2 $\Omega$. Going round the loop:

The voltage drops are 8 V and 4 V, and they add up to the 12 V of the source.

The capacitor is the only element that stores charge here. In DC steady state it acts as an open circuit, and it can hold its voltage when the switch is closed. Take the 5 V source with its + terminal at the top, and the bottom wire as the reference.

The capacitor is charged to 5 V.

Check: the currents into the node are $0.2$ A and $0.1$ A, and the current to ground is $3/10 = 0.3$ A.

The dual of a network is found by swapping each quantity with its dual:

• a mesh becomes a node, and series elements become parallel elements,
• $L \leftrightarrow C$, $R \leftrightarrow G$ (1 $\Omega$ becomes 1 S), voltage source $\leftrightarrow$ current source.

Mesh 1 holds 5 V, 2 H and 1 F. Mesh 2 holds 1 F, 4 H and 2 $\Omega$. The 1 A source is common. In the dual they become two nodes and a reference node:

• Node 1 to reference: 5 A source, 2 F and 1 H, in parallel.
• Node 1 to node 2: a 1 V source.
• Node 2 to reference: 1 H, 4 F and 2 S, in parallel.

When the switch $k$ opens at $t = 0$, the 2 A source feeds the parallel combination of the 0.5 $\Omega$ resistor, the 0.5 H inductor and the 1 F capacitor. Both $L$ and $C$ start from zero.

Check: $v(0) = 0$ as the capacitor has no initial charge. Also $\dfrac{dv}{dt}(0) = 2$ V/s, and $\dfrac{i_C}{C} = \dfrac{2}{1}$ since at $t = 0^+$ the full 2 A flows in the capacitor.

Reciprocity means the ratio of the response to the excitation stays the same when the source and the response are interchanged. The g parameters are $I_1 = g_{11}V_1 + g_{12}I_2$ and $V_2 = g_{21}V_1 + g_{22}I_2$.

• Case A: apply a voltage $V_a$ at port 1, with port 2 open ($I_2 = 0$). Then $V_2 = g_{21}V_a$.
• Case B: apply a current $I_b$ at port 2, with port 1 shorted ($V_1 = 0$). Then $I_1 = g_{12}I_b$.

For a reciprocal network the interchange rule is $V_{1a}I_{1b} + V_{2a}I_{2b} = V_{1b}I_{1a} + V_{2b}I_{2a}$. Here $V_{1b} = 0$ and $I_{2a} = 0$, so the right side is zero, and $V_a(g_{12}I_b) + (g_{21}V_a)I_b = 0$. Hence

The z parameters are $V_1 = z_{11}I_1 + z_{12}I_2$ and $V_2 = z_{21}I_1 + z_{22}I_2$. The h parameters are $V_1 = h_{11}I_1 + h_{12}V_2$ and $I_2 = h_{21}I_1 + h_{22}V_2$. In both, $I_1$ is independent, but the h set takes $V_2$ in place of $I_2$. So to go from z to h, we only need to swap $I_2$ for $V_2$ as the independent variable. This is a simple algebra step with two equations.

Step 1: from the second z equation, solve for $I_2$:

Comparing with the h equation, $h_{21} = -\dfrac{z_{21}}{z_{22}}$ and $h_{22} = \dfrac{1}{z_{22}}$.

Step 2: put this $I_2$ in the first z equation:

where $\Delta z = z_{11}z_{22} - z_{12}z_{21}$.

The h parameters are very useful for transistor circuits, since $h_{21}$ is the current gain. Check: if $z_{12} = z_{21}$ (reciprocal), then $h_{12} = -h_{21}$ follows at once from these four results, which is the known condition for the h parameters.

After the switch is closed, the charged capacitor ($V_0 = 10$ V, positive on the top plate) discharges through the inductor. Take $i$ as shown, leaving the positive plate. Here $C = 2$ F and $L = 2$ H.

Check: $i(0) = 0$, and $\dfrac{di}{dt}(0) = 5 = \dfrac{V_0}{L}$, since the full 10 V is across the 2 H inductor at $t = 0^+$.