Mastering Circuits!
Mastering the Circuit Theorems in Electrical Circuits.
Linearity Property
Linearity: Linearity is the property of an element describing a linear relationship between cause and effect.
This property is a combination of both the homogeneity (scaling) property and the additivity property.
Homogeneity Property: The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called the response) is multiplied by the same constant.
\small \begin{aligned} &\text{Ohm’s Law relates the input } i \text{ to the output } v \\[0.5ex] & \text{ as,} \\[2ex] & \quad v = iR \\[1ex] & \quad \therefore kv = kiR \quad [k = \text{constant} ] \\[2ex] &\text{Suppose,} \\[1ex] & \quad v = 10 \ V \text{ gives } i = 2 \ A \\[2ex] &\text{Then,} \\[1ex] & \quad v = 1 \ V \text{ will give } i = 0.2 \ A \ [\text{multiplied by } 0.1] \\[1ex] & \quad v = 20 \ V \text{ will give } i = 4 \ A \ \ [\text{multiplied by } 2] \end{aligned}
Additivity Property: The additivity property requires that the response to a sum of inputs is the sum of responses to each input applied separately.
\small \begin{aligned} &\text{If } v_1 = i_1R \text{ and } v_2 = i_2R \\[1ex] &\text{then applying } (i_1 + i_2) \text{ gives,} \\[1ex] & v = (i_1 + i_2)R = i_1R + i_2R = v_1 + v_2 \\[1ex] & \therefore v = v_1 + v_2 \end{aligned}
Linear Circuit: A linear circuit is one whose output is linearly related (or directly proportional) to its input.
A linear circuit consists of only linear elements, linear dependent sources and independent sources.
Figure 1 – A linear circuit with input v_s and output i
Since p = i^{2}R = \frac{v^{2}}{R} (making it a quadratic function rather than a linear one), the relationship between power and voltage (or current) is non-linear.
\small \begin{aligned} &\text{If } p_1 = i_1^2R \text{ and } p_2 = i_2^2R \\[1ex] &\text{then applying } (i_1 + i_2) \text{ gives,} \\[1ex] & p = (i_1 + i_2)^2R = i_1^2R + 2i_1i_2R + i_2^2R \\[1ex] & \therefore p \neq p_1 + p_2 \end{aligned}
As it doesn’t follow the additivity property, therefore the relationship between power and voltage (or current) is non-linear.
Solved Problems on Linearity Property
Problem 1
Pb-1: For the circuit in Fig. 2, find I_{0} when v_{s} = 24 \ V and v_{s} = 12 \ V .
Figure 2 – Circuit diagram for Pb-1
Solution
Figure 3 – Circuit diagram for solving the Pb-1
\small \begin{aligned} &\text{KVL at mesh 1,} \\[2ex] & 6i_1 + 2i_1 + 4(i_1 - i_2) + v_s = 0 \\[2ex] & \Rightarrow 6i_1 + 2i_1 + 4i_1 - 4i_2 = -v_s \\[2ex] & \therefore 12i_1 - 4i_2 = -v_s \quad \text{\text{-} \text{-} \text{-} (i)} \\[3ex] &\text{KVL at mesh 2,} \\[2ex] & -v_s + 4(i_2 - i_1) + 8i_2 + 4i_2 - 3v_x = 0 \\[2ex] & \Rightarrow 4i_2 - 4i_1 + 8i_2 + 4i_2 - 3 \cdot 2i_1 = v_s \\[2ex] & \Rightarrow 4i_2 - 4i_1 + 8i_2 + 4i_2 - 6i_1 = v_s \\[2ex] & \therefore -10i_1 + 16i_2 = v_s \quad \text{\text{-} \text{-} \text{-} (ii)} \\[3ex] &\text{When } v_s = 24 \ V, \\[1ex] & \quad \text{From (i),} \\[1ex] & \quad \quad 12i_1 - 4i_2 = -24 \quad \text{\text{-} \text{-} \text{-} (iii)} \\[1ex] & \quad \text{From (ii),} \\[1ex] & \quad \quad -10i_1 + 16i_2 = 24 \quad \text{\text{-} \text{-} \text{-} (iv)} \\[1ex] & \quad \text{Solving equations (iii) \& (iv),} \\[1ex] & \quad \quad \therefore i_1 = -\frac{36}{19} \ A, \ i_2 = \frac{6}{19} \ A \\[1.5ex] & \quad \quad \therefore I_0 = i_2 = \frac{6}{19} \ A \\[3ex] &\text{When } v_s = 12 \ V, \\[1ex] & \quad \text{From (i),} \\[1ex] & \quad \quad 12i_1 - 4i_2 = -12 \quad \text{\text{-} \text{-} \text{-} (v)} \\[1ex] & \quad \text{From (ii),} \\[1ex] & \quad \quad -10i_1 + 16i_2 = 12 \quad \text{\text{-} \text{-} \text{-} (vi)} \\[1ex] & \quad \text{Solving equations (v) \& (vi),} \\[1ex] & \quad \quad \therefore i_1 = -\frac{18}{19} \ A, \ i_2 = \frac{3}{19} \ A \\[1.5ex] & \quad \quad \therefore I_0 = i_2 = \frac{3}{19} \ A \end{aligned}
Problem 2
Pb-2: For the circuit in Fig. 4, find v_{0} when i_{s} = 30 \ A and i_{s} = 45 \ A .
Figure 4 – Circuit diagram for Pb-2
Solution
Figure 5 – Circuit diagram for solving the Pb-2
\small \begin{aligned} &\text{KCL at node 1,} \\[2ex] & \quad -i_s + \frac{v_1}{4} + \frac{v_1-v_2}{12} = 0 \\[2ex] & \quad \Rightarrow \frac{3v_1+v_1-v_2}{12} = i_s \\[2ex] & \quad \therefore 4v_1 - v_2 = 12i_s \quad \text{\text{-} \text{-} \text{-} (i)} \\[3ex] &\text{KCL at node 2,} \\[2ex] & \quad \frac{v_2-v_1}{12} + \frac{v_2}{8} = 0 \\[2ex] & \quad \Rightarrow \frac{2v_2-2v_1+3v_2}{24} = 0 \\[2ex] & \quad \therefore -2v_1 + 5v_2 = 0 \quad \text{\text{-} \text{-} \text{-} (ii)} \\[3ex] &\text{When } i_s = 30 \ A, \\[1ex] & \quad \text{From (i),} \\[1ex] & \quad \quad 4v_1 - v_2 = 12 \times 30 \\[1ex] & \quad \quad \therefore 4v_1 - v_2 = 360 \quad \text{\text{-} \text{-} \text{-} (iii)} \\[1ex] & \quad \text{Solving equations (ii) \& (iii),} \\[1ex] & \quad \quad \therefore v_1 = 100 \ V, \ v_2 = 40 \ V \\[1.5ex] & \quad \quad \therefore v_0 = v_2 = 40 \ V \\[3ex] &\text{When } i_s = 45 \ A, \\[1ex] & \quad \text{From (i),} \\[1ex] & \quad \quad 4v_1 - v_2 = 12 \times 45 \\[1ex] & \quad \quad \therefore 4v_1 - v_2 = 540 \quad \text{\text{-} \text{-} \text{-} (iv)} \\[1ex] & \quad \text{Solving equations (ii) \& (iv),} \\[1ex] & \quad \quad \therefore v_1 = 150 \ V, \ v_2 = 60 \ V \\[1.5ex] & \quad \quad \therefore v_0 = v_2 = 60 \ V \end{aligned}
Problem 3
Pb-3: Assume I_{0} = 1 \ A and use linearity to find the actual value of I_{0} in the circuit of Fig. 6.
Figure 6 – Circuit diagram for Pb-3
Solution
Figure 7 – Circuit diagram for solving the Pb-3
\small \begin{aligned} &\text{Assuming } I_0 = 1 \ A, \\[2ex] & \quad \therefore V_1 = (3 + 5)I_0 = 8 \times 1 = 8 \ V \\[2ex] & \quad \& \ I_1 = \frac{V_1}{4} = \frac{8}{4} = 2 \ A \\[2ex] & \quad \text{KCL at node 1,} \\[2ex] & \quad I_0 + I_1 - I_2 = 0 \Rightarrow 1 + 2 - I_2 = 0 \\[2ex] & \quad \therefore I_2 = 3 \ A \\[3ex] & \quad \text{Now,} \\[2ex] & \quad \quad I_2 = \frac{V_2 - V_1}{2} \\[2.5ex] & \quad \quad \Rightarrow 3 = \frac{V_2 - 8}{2} \\[2ex] & \quad \quad \therefore V_2 = 14 \ V \\[3ex] & \quad \text{Again,} \\[2ex] & \quad \quad I_3 = \frac{V_2}{7} = \frac{14}{7} = 2 \ A \\[3ex] & \quad \text{KCL at node 2,} \\[2ex] & \quad I_2 + I_3 - I_4 = 0 \Rightarrow 3 + 2 - I_4 = 0 \\[2ex] & \quad \therefore I_4 = 5 \ A \\[2ex] &\therefore I_s = I_4 = 5 \ A \\[3ex] &\text{This shows that assuming } I_0 = 1 \ A \text{ gives } \\[0.5ex] & I_s = 5 \ A. \text{ Therefore, the actual source current, } \\[0.5ex] & I_s = 15 \ A \text{ will give } I_0 = 3 \ A \text{ as the actual value.} \end{aligned}
Problem 4
Pb-4: Assume that V_{0} = 1 \ V and use linearity to calculate the actual value of V_{0} in the circuit of Fig. 8.
Figure 8 – Circuit diagram for Pb-4
Solution
Figure 9 – Circuit diagram for solving the Pb-4
\small \begin{aligned} &\text{Assuming } V_0 = 1 \ V, \\[2ex] & \quad \therefore I_1 = \frac{V_0}{8} = \frac{1}{8} \ A \\[3ex] & \quad \text{Now,} \\[2ex] & \quad I_1 = \frac{V_1 - V_0}{12} \\[2ex] & \quad \Rightarrow \frac{1}{8} = \frac{V_1 - 1}{12} \\[2ex] & \quad \therefore V_1 = 2.5 \ V \\[2ex] &\therefore V_s = V_1 = 2.5 \ V \\[3ex] &\text{This shows that assuming } V_0 = 1 \ V \text{ gives } \\[0.5ex] & V_s = 2.5 \ V. \text{ Therefore, the actual source } \\[0.5ex] & \text{ voltage, } V_s = 40 \ V \text{ will give } V_0 = 16 \ V \\[0.5ex] & \text{ as the actual value. } \end{aligned}
SuperPosition Theorem
The superposition principle states that the voltages across (or currents through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
It helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately.
Fundamental Rules of Superposition
To apply the superposition principle correctly, we must keep these two rules in mind
1. Isolate Independent Sources
Consider only one independent source at a time. All other independent sources must be “turned off”.
- Voltage Sources are replaced by 0 \ V (or a short circuit)
- Current Sources are replaced by 0 \ A (or an open circuit)
Let’s suppose that we need to find v_{0} using SuperPosition Principle in the circuit of Fig. 10.
Figure 10 – Circuit diagram to apply superposition principle
- Voltage source will be short circuited.
Figure 11 – Short circuited voltage source for superposition principle
- Current source will be open circuited.
Figure 12 – Open circuited current source for superposition principle
2. Leave Dependent Sources Intact
Never turn off dependent sources because they are controlled by circuit variables.
Let’s suppose that we need to find i_{0} and v_{0} using SuperPosition Principle in the circuit of Fig. 13.
Figure 13 – Circuit diagram to apply superposition principle (with dependent source)
- Dependent source won’t be replaced.
Figure 14 – Short circuited voltage source for superposition principle (with dependent source)
Figure 15 – Open circuited current source for superposition principle (with dependent source)
Steps to Apply Superposition Theorem
Figure 16 – Circuit diagram to apply superposition theorem
we need to find i_{0} and v_{0} using SuperPosition Theorem in the circuit of Fig. 16.
1. Activate One Source at a Time
Select one independent source to be active. Deactivate all other independent sources (short circuit the voltage sources and open circuit the current sources).
Figure 17 – Activating current source
Figure 18 – Activating voltage source
2. Solve for All Independent Sources
Find the required output (the specific voltage or current that are asked to find) due to that single active source. Use standard techniques like Nodal Analysis, Mesh Analysis, Ohm’s Law, and etc. Repeat it for all independent sources.
\small \begin{aligned} &\text{Let,} \\[1ex] & \quad v_0 = v_0' + v_0'' \\[2ex] &\text{Where,} \\[1ex] & \quad v_0' = \text{contribution due to } 6 \ A \text{ current source} \\[1ex] & \quad v_0'' = \text{contribution due to } 30 \ V \text{ voltage source} \end{aligned}
Figure 19 – Solving for current source
\small \begin{aligned} &\text{KCL at node 1,} \\[2ex] & -6 + \frac{v_1}{40} + \frac{v_1-v_2}{10} = 0 \\[2ex] & \Rightarrow \frac{-240+v_1+4v_1-4v_2}{40} = 0 \\[2ex] & \therefore 5v_1 - 4v_2 = 240 \quad \text{\text{-} \text{-} \text{-} (i)} \\[3ex] &\text{KCL at node 2,} \\[2ex] & \frac{v_2}{20} - 4i_0' + \frac{v_2-v_1}{10} = 0 \\[2ex] & \Rightarrow \frac{v_2}{20} - 4 \cdot \frac{v_1-v_2}{10} + \frac{v_2-v_1}{10} = 0 \\[2.5ex] & \Rightarrow \frac{v_2-8v_1+8v_2+2v_2-2v_1}{20} = 0 \\[2ex] & \therefore -10v_1 + 11v_2 = 0 \quad \text{\text{-} \text{-} \text{-} (ii)} \\[3ex] &\text{Solving equations (i) and (ii),} \\[2ex] & \therefore v_1 = 176 \ V, \ v_2 = 160 \ V \\[2ex] & \therefore v_0' = v_1 - v_2 = 176 - 160 = 16 \ V \end{aligned}
Figure 20 – Solving for voltage source
\small \begin{aligned} &\text{KCL at node 3,} \\[2ex] & \frac{v_3}{40} + \frac{v_3-v_4}{10} = 0 \\[2ex] & \Rightarrow \frac{v_3+4v_3-4v_4}{40} = 0 \\[2ex] & \therefore 5v_3 - 4v_4 = 0 \quad \text{---(iii)} \\[3ex] &\text{KCL at node 4,} \\[2ex] & \frac{v_4-(-30)}{20} - 4i_0'' + \frac{v_4-v_3}{10} = 0 \\[2ex] & \Rightarrow \frac{v_4+30}{20} - 4 \cdot \frac{v_3-v_4}{10} + \frac{v_4-v_3}{10} = 0 \\[2.5ex] & \Rightarrow \frac{v_4+30-8v_3+8v_4+2v_4-2v_3}{20} = 0 \\[2ex] & \therefore -10v_3 + 11v_4 = -30 \quad \text{---(iv)} \\[3ex] &\text{Solving equations (iii) and (iv),} \\[2ex] & \therefore v_3 = -8 \ V, \ v_4 = -10 \ V \\[2ex] & \therefore v_0'' = v_3 - v_4 = -8 - (-10) = 2 \ V \end{aligned}
3. Algebraic Summation
Find the total value of the desired variable by taking the algebraic sum of all individual contributions calculated in the previous steps.
\small \begin{aligned} &\text{We get,} \\[2ex] & \quad v_0' = 16 \ V \\[2ex] & \quad v_0'' = 2 \ V \\[3ex] &\text{Now,} \\[2ex] & \quad v_0 = v_0' + v_0'' = 16 + 2 \\[2ex] & \quad \therefore v_0 = 18 \ V \\[3ex] &\text{Again,} \\[2ex] & \quad i_0 = \frac{v_0}{10} = \frac{18}{10} \\[3ex] & \quad \therefore i_0 = 1.8 \ A \end{aligned}