# Category Archives: Electrical Circuits Problems

All Solved Problems in Electrical Circuits

# Total Energy Stored - Circuit with Capacitors and Inductors

Find the total energy stored in the circuit.

Fig. (1-28-1) - The circuit

Solution
The circuit contains only dc sources. Recall that an inductor is a short circuit to dc and a capacitor is an open circuit to dc. These can be easily verified from their current-voltage characteristics. For an inductor, we have $v(t)=L \frac{d i(t)}{dt}$. Since a dc current does not vary with time, $\frac{d i(t)}{dt}=0$. Hence, the voltage across the inductor is zero for any dc current. This is to say that dc current passes through the inductor without any voltage drop, exactly similar to a short circuit. For a capacitor, the current-voltage terminal characteristics is $i(t)=L \frac{d v(t)}{dt}$. Voltage drop across passive elements due to dc currents does not vary with time. Therefore, $\frac{d v(t)}{dt}=0$ and consequently the current of the capacitor is zero. This is to say that dc current does not pass through the capacitor regardless of the voltage amount. This is similar to the behavior of an open circuit. Please note that unlike dc current, ac current passes through capacitors in general.

# Thévenin's Theorem - Circuit with Two Independent Sources

Use Thévenin's theorem to determine $I_O$.

Fig. (1-27-1) - Circuit with two independent sources

Solution
Lets break the circuit at the $3\Omega$ load as shown in Fig. (1-27-2).

# Thévenin's Theorem - Circuit with An Independent Source

Use Thévenin's theorem to determine $V_O$.

Fig. (1-26-1) - The Circuit

Solution
To find the Thévenin equivalent, we break the circuit at the $4\Omega$ load as shown below.

# Superposition Method - Circuit With Dependent Sources

Determine $I_x$, $I_y$ and $V_z$ using the superposition method.

Solution
I. Contribution of the $-2V$ voltage source:
We need to turn off the current source by replacing it with an open circuit. Recall that we do not turn off dependent sources. The resulting circuit is shown below.

# Superposition Problem with Four Voltage and Current Sources

Determine $V_x$ and $I_x$ using the superposition method.

Solution
I. Contribution of the $-5V$ voltage source:

To find the contribution of the $-5V$ voltage source, other three sources should be turned off. The $3V$ voltage source should be replaced by short circuit. The current source should be replaced with open circuits, as shown below.

# Nodal Analysis - Circuit with Dependent Voltage Source

Determine the power of each source after solving the circuit by the nodal analysis.

Answers: $P_{I_x}=0.497W, P_{1A}=-1.806W, P_{2A}=4.254W, P_{3V}=-3.87W,$ and $P_{5V}=-3.552W$

Solution

I. Identify all nodes in the circuit.
The circuit has 6 nodes as highlighted below.

# Nodal Analysis – 6-Node Circuit

Determine the power of each source after solving the circuit by the nodal analysis.

Solution
I. Identify all nodes in the circuit.
The circuit has 6 nodes as indicated below.

# Nodal Analysis - Dependent Voltage Source

Use nodal analysis method to solve the circuit and find the power of the $3\Omega$- resistor.

Solution

I. Identify all nodes in the circuit.
The circuit has 3 nodes as shown below.

# Nodal Analysis - Dependent Current Source

Deploy nodal analysis method to solve the circuit and find the power of the dependent source.

Solution
I. Identify all nodes in the circuit. Call the number of nodes $N$.
The circuit has 4 nodes:

Therefore, $N=4$.
Solve the circuit with the nodal analysis and determine $I_x$.
1) Identify all nodes in the circuit. Call the number of nodes $N$.