Solved Problems on Thevenin's Theorem
Demonstrative Video
Problem-1
Find the load current \(i\) using the Thevenin’s Theorem?
Solution-1
Applying mesh analysis \[\begin{aligned} 3i_1-i_2&=10\\ -i_1+4i_2&=-5 \end{aligned}\]
On solving: \[\begin{aligned} i_1&=5/11, \quad i_2 = -5/11 \end{aligned}\]
Thevenin or \(V_{oc}\) \[\begin{aligned} V_{TH}&=(5+2i_2)=(5-\dfrac{10}{11})=\dfrac{45}{11}~\mathrm{V} \end{aligned}\]
Equivalent resistance, \(R_{\text {TH }}=\dfrac{\dfrac{5}{3} \times 2}{5 / 3+2}=\dfrac{10}{11} \Omega\)
Load current is, \(i=\dfrac{V_{o c}}{R_{\mathrm{TH}}+2}=\dfrac{45 / 11}{10 / 11}=\dfrac{45}{32}=1.40625 \mathrm{~A}\)
Problem-2
Find the current through 15 \(\Omega\) using the Thevenin’s Theorem?
Solution-2
Applying KVL \(20 I+5 I+10-40=0\) \[\begin{gathered}
25 I=30 \\
I=1.2 \mathrm{~A} \\
V_{T H}=10+\left(5\times1.2\right)=16 v
\end{gathered}\]
\[\begin{aligned}
&R_{T H}=\frac{5 \times 20}{5+20}=4 \Omega \\
&I_{L}=\frac{V_{T H}}{R_{T
H}+R_{L}}=\frac{16}{4+15}=0.842 A
\end{aligned}\]
Problem-3
Find the Thevenin’s volatge and resistance at terminals 1-2?
Solution-3
SC the voltage source and OC the current source \(R_{TH}=4\Omega\)
Thevenin’s voltage \(V_{TH}=4\mathrm{V}\)
Problem-4
Find the Thevenin’s resistance at terminals P-Q?
Solution-4
Connecting 1V source by SC indep. voltage source \(I=1\mathrm{A}\)
\(R_{TH}=V/I=1\Omega\)
Problem-5
Find the current flowing through A-B using Thevenin’s theorem?
Solution-5
Open-circuiting the \(4-\Omega\) resistor by \(\mathrm{KCL}\), \[\frac{V_{\mathrm{oc}}-10}{2}=4 v_{s}=4\left(10-V_{\mathrm{oc}}\right) \Rightarrow V_{\mathrm{oc}}=10 \mathrm{~V}\]
SC the terminals AB \[\begin{aligned} &\frac{V_{1}-10}{2}+\frac{V_{1}}{4}=4 v_{s}=4\left(10-V_{1}\right) \\ &\quad V_{1}=\frac{180}{19}=9 \cdot 47 \mathrm{~V} \\ &\therefore \quad I_{\mathrm{sc}}=\frac{9 \cdot 47}{4}=2 \cdot 368 \mathrm{~A} \\ &\therefore \quad R_{\mathrm{th}}=\frac{V_{\mathrm{th}}}{I_{\mathrm{sc}}}=4 \cdot 22 \Omega \end{aligned}\]
Problem-6
Find the Thevenin’s equivalence at terminals a-b?
Solution-6
By KVL for LHS loop, \[1 \times 10^{3} \times I+\frac{V_{0}}{10^{4}}=10 \times 10^{-3}\]
In the RHS loop, the dependent current source will circulate in the resistor. By KVL, \[V_{0}=30 \times 10^{3} \times(-75 I)=-225 \times 10^{4} I\]
Substituting the value of \(I\) from (ii) in (i), we get, \[\begin{aligned} &\Rightarrow 1 \times 10^{3} \times\left(-\frac{V_{0}}{225 \times 10^{4}}\right)+\frac{V_{0}}{10^{4}}=10 \times 10^{-3} \\ &\Rightarrow-4.44 \times 10^{-4} V_{0}+1 \times 10^{-4} V_{0}=10 \times 10^{-3} \\ &\Rightarrow V_{0}=-\frac{10 \times 10^{-3}}{3.44 \times 10^{-4}}=-29 \mathrm{~V} \end{aligned}\]
Now, SC the terminals \(A\) and \(B\), we get by KVL to LHS loop, \[1 \times 10^{3} \times I+0=10 \times 10^{-3} \Rightarrow I=1 \times 10^{-5} \mathrm{~A}\]
Also, from RHS loop on the short circuit, \(I_{\mathrm{sc}}=-75 I=-75 \times 1 \times 10^{-5}=-75 \times 10^{-5} \mathrm{~A}\)
Thus, the Thevenin equivalent impedance is given as \[Z_{\mathrm{th}}=\frac{V_{\mathrm{oc}}}{I_{\mathrm{sc}}}=\frac{-29}{-75 \times 10^{-5}}=38.67 \mathrm{k} \Omega\]
Problem-7
Find the Thevenin’s resistance at terminals C-D?
Solution-7
SC the voltage source and connect \(V_x=1\mathrm{V}\)
\(R_{TH}=V_x/I_x = 20~k\Omega\)
Problem-8
Find the Thevenin’s equivalence at terminals a-b?
Solution-8
\(I_{sc}=100/20=5\Omega\)
Applying KCL in open circuit \[-\frac{v_{oc}}{100}+\frac{100+v_{oc}}{20}=0 \Rightarrow-v_{oc}+500+5 v_{o c}=0 \Rightarrow v_{o c}=-125 \mathrm{~V}\]
\(R_{TH}=\dfrac{V_{oc}}{I_{sc}}= 25\Omega\)