Solved Problems on Superposition Theorem
Demonstrative Video
Problem-1
Find the current \(I\) using Superposition theorem?
Solution-1
When voltage source acting alone \(I^{'} = -1/3\) A
When current source is acting alone
\[I^{''} = 1\times
\dfrac{1}{1+2}=1/3~\mathrm{A}\]By Superposition: \[I=(I^{'}+I^{''}) = -1/3+1/3=0\]
Problem-2
Use superposition theorem to determine current \(I\) ?
Solution-2
When 10-V voltage source is acting alone
\[\begin{aligned} \text{KVL} \Rightarrow 5i^{'}-2v^{'}_x+2i^{'}& = 10; \quad v^{'}_x = -2i^{'} \\ \Rightarrow 7i^{'}+4i^{'} & =10 \Rightarrow i^{'} = 10/11~\mathrm{A} \end{aligned}\]
When 2-A current source is acting alone
\[\begin{aligned}
&\text{KCL} \Rightarrow 2 = i_x +
i^{''} = -\dfrac{v^{'}_x}{2}+i^{''} \\
&\text{KVL} \Rightarrow 5i^{''} +
3v^{''}_x = 0\\
& \quad \Rightarrow i^{''} =
-\dfrac{3}{5}v^{''}_x
\end{aligned}\]
\[\begin{gathered} 2=-\frac{v_{x}^{\prime \prime}}{2}-\frac{3}{5} v_{x}^{\prime \prime} \Rightarrow v_{x}^{\prime \prime}=-\frac{20}{11} \\ \therefore i^{\prime \prime}=-\frac{3}{5} \times\left(-\frac{20}{11}\right)=\frac{12}{11} \mathrm{~A} \\ I=\left(i^{\prime}-i^{\prime \prime}\right)=\left(\frac{10}{11}-\frac{12}{11}\right)=-\frac{2}{11} \mathrm{~A} \end{gathered}\]
Problem-3
Find \(i_0\) and \(i\) using Superposition ?
Solution-3
When 6-V source acting alone
\[\begin{aligned} i^{'}_0 & = i^{'} \\ \text{KVL} \Rightarrow & 6i^{'}+2i^{'}=6\\ \Rightarrow & i^{'} = i^{'}_0=0.75~\mathrm{A} \end{aligned}\]
When 1-A current source is acting alone
\[\begin{aligned} \text{KCL} \Rightarrow 1&=i^{\prime \prime}-i_{0}^{\prime \prime} \Rightarrow i^{\prime \prime}=1+i_{0}^{\prime \prime} \\ \text{KVL in supermesh} \Rightarrow & 1 \times i_{0}^{\prime \prime}+5 i^{\prime \prime}+2 i_{0}^{\prime \prime}=0\\ & 3 i_{0}^{\prime \prime}+5 i^{\prime \prime}=0\\ & 3 i_{0}^{\prime \prime}+5\left(1+i_{0}^{\prime \prime}\right)=0\\ & i_{0}^{\prime \prime}=-\frac{5}{4}=-1.25 \mathrm{~A}\\ \therefore ~ & i^{\prime \prime}=1-1.25=-0.25 \mathrm{~A} \end{aligned}\]
By superposition:
\[\left.\begin{array}{l} i=\left(i^{\prime}+i^{\prime \prime}\right)=(0.75-0.25)=0.5 \mathrm{~A} \\ i_{0}=\left(i_{0}^{\prime}+i_{0}^{\prime \prime}\right)=(0.75-1.25)=-0.5 \mathrm{~A} \end{array}\right\}\]
Problem-4
Find \(V_{AB}\) using Superposition theorem ?
Solution-4
When 2-V source acting alone
\[\begin{aligned}
I^{\prime} & = 2/12 = 1/6~\mathrm{A} \\
V^{\prime}_{AB} & = I^{\prime} \times 6 =
1~\mathrm{V}
\end{aligned}\]
When 4-V source acting alone
\[\begin{aligned}
I^{''} & = 4/12 = 1/3~\mathrm{A} \\
V^{''}_{AB} & = I^{''} \times 6 =
-2~\mathrm{V}
\end{aligned}\]When 2-A source acting alone, convert into equivalent voltage source
\[\begin{aligned}
I^{'''} & = 8/12 = 2/3~\mathrm{A} \\
V^{'''}_{AB} & = - I^{'''}
\times 6 = -4~\mathrm{V}
\end{aligned}\]By Superposition \[V_{A B}=V_{A B}^{\prime}+V_{A B}^{\prime \prime}+V_{A B}^{\prime \prime \prime}=(1-2-4)=-5 \mathrm{~V}\]
Problem-5
Determine \(V_1\) using Superposition theorem ?
Solution-5
When 8-A source is acting alone
\[\begin{aligned} \text{KVL in supermesh} \Rightarrow & 3 i^{\prime}+2 i_{1}-4 i^{\prime}=0 \Rightarrow i_{1}=\frac{1}{2} i^{\prime}\\ \text{KCL at node} x \Rightarrow & i_{1}=\left(8+i^{\prime}\right) \Rightarrow \frac{1}{2} i^{\prime}=8+i^{\prime} \Rightarrow i^{\prime}=-16 \mathrm{~A} \\ \therefore ~ & V_{1}^{\prime}=3 i^{\prime}=3 \times(-16)=-48 \mathrm{~V} \end{aligned}\]
When 2-A source is acting alone
\[\begin{aligned} \text{KVL} \Rightarrow & 3\left(i_{2}+2\right)+2 i_{2}-4 i^{\prime \prime}=0 \Rightarrow 5 i_{2}+6-4 i^{\prime \prime}=0 \\ i^{\prime \prime}&=\left(i_{2}+2\right) \\ &\therefore 5 i_{2}+6-4\left(i_{2}+2\right)=0 \Rightarrow i_{2}=2 \mathrm{~A} \\ &\therefore i^{\prime \prime}=\left(i_{2}+2\right)=(2+2)=4 \mathrm{~A} \\ &\therefore V_{1}^{\prime \prime}=3 i^{\prime \prime}=3 \times 4=12 \mathrm{~V} \end{aligned}\]
When 10-V source is acting alone
By KVL, \[\begin{aligned} &3 i^{\prime \prime \prime}-10+2 i^{\prime \prime \prime}-4 i^{\prime \prime \prime}=0 \Rightarrow i^{\prime \prime \prime}=10 \mathrm{~A} \\ \therefore & V_{1}^{\prime \prime \prime}=10 \times 3=30 \mathrm{~V} \end{aligned}\]
When all the sources are acting simultaneously, by the superposition theorem the voltage is given as \[V_{1}=\left(V_{1}^{\prime}+V_{1}^{\prime \prime}+V_{1}^{\prime \prime \prime}\right)=(-48+12+30)=-6 \mathrm{~V}\]
Problem-6
Find \(I\) using superposition theorem ?
Solution-6
4-V source alone
\[\begin{aligned} \text{KVL} ~ -4+3 I^{\prime}+5 V_{x}^{\prime}-V_{x}^{\prime} &=0 \\ 3 I^{\prime}+4 \times\left(-2 I^{\prime}\right) &=4 \\ \left[\because V_{x}^{\prime}=-2 I^{\prime}\right] & \end{aligned}\]
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2-A source alone
\[\begin{aligned} \text {KCL, } & 2=\frac{V_{x}^{\prime \prime}}{2}+\frac{V_{x}^{\prime \prime}-5 V_{x}^{\prime \prime}}{3} \\ & \Rightarrow V_{x}^{\prime \prime}=-\frac{12}{5}=-2.4 \mathrm{~V} \\ \therefore & I^{\prime \prime}=\frac{V_{x}^{\prime \prime}-5 V_{x}^{\prime \prime}}{3} =3.2 \mathrm{~A} \end{aligned}\]
By superposition theorem \[I=\left(I^{\prime}+I^{\prime \prime}\right)=(-0.8+3.2)=2.4 \mathrm{~A}\]