Solved Problems on Norton's Theorem
Demonstrative Video
Problem-1
Determine Norton equivalent current and resistance at terminal a-b?
Solution-1
Applying KCL at node \(a\) \[\begin{aligned} I_{N}+I_{1}+2&=0\\ I_{N}-\frac{5}{6}+2 &=0 \\ I_{N} &=-\frac{7}{6} \mathrm{~A} \end{aligned}\]
\[R_N=24~\Omega\]
Problem-2
Determine \(V_0\) using Norton’s theorem?
Solution-2
Norton resistance:
\[R_{N} = 10+2+(12||24) = 20~{k\Omega}\]
Norton current:
Apply superposition theorem by keeping one source active at a time \[I_{N} = I_{v}+I_{i}\]
Considering voltage source only
\[I_{1} = \dfrac{8}{8+12}\times 3 = 1.2~\mbox{mA}
\quad 12 || 24 = 8~{k\Omega}\]Considering current source only, we will get parallel equivalent resistance \[I_{2} = -1.5~\mbox{mA}\]
Norton current \(I_{N}=1.2-1.5=-0.3~\mbox{mA}\)
Norton equivalent circuit
\[V_{0} = (1~{k\Omega})\dfrac{20}{20+1}\times
(-0.3~\mbox{mA}) = -0.285~\mbox{V}\]
Problem-3
Determine Norton equivalent current and resistance at terminal a-b?
Solution-3
Determine Norton resistance
\[R_{N} = 12 || (20+40) =
10~\Omega\]
Determine Norton current
Appling source transformation and KVL
\[\begin{aligned}
-120+80+60I_{N}&=0\\
I_{N}&=0.66~A
\end{aligned}\]
Problem-4
Determine the current \(i\) using Norton equivalent circuit?
Solution-4
Norton resistance and current
\[\begin{aligned}
R_{N} & = 6+4=10~\Omega\\
2+(12-v)/6&=v/4 \Rightarrow v = 9.6~V\\
I_{N}&=-(12-v)/6=0.4 \Rightarrow I_{N} =
-0.4~A
\end{aligned}\]
Combining the RHS circuit to Norton equivalent
\[\begin{aligned}
i & = \dfrac{10}{10+5}(4-0.4)=2.4~A
\end{aligned}\]
Problem-5
Find the Norton equivalent circuit ?
Solution-5
By KCL at node \(v_1\) \[\begin{aligned} \dfrac{v_1}{8}+\dfrac{v_1}{2}+\dfrac{v_1-v_2}{2}&=0 \end{aligned}\]
By KCL at node \(v_2\) \[\begin{aligned} i_1+2i_1&=\dfrac{v_2}{2} \end{aligned}\]
Also, \[i_1=\dfrac{v_1-v_2}{2}\]
On solving \[\begin{aligned} v_1&=0~V\\ v_2&=0~V\\ i_{sc} &=\dfrac{v_1}{2}-2i_1=\dfrac{v_1}{2}-2(\dfrac{v_1-v_2}{2})=0~A \end{aligned}\]
\[R_0 = \dfrac{v_0}{i_0}=3~\Omega\]