With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed: i) 1 Ω connected at port B draws a current of 3 A ii) 2.5 Ω connected at port B draws a current of 2 A
With 6 V dc connected at port A, 1 Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is
8 V
Now, when 6 V connected at port A let thevenin voltage seen at port B is Vth,6 V . Here RL = 1 Ω and IL = 7/3 A
\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSaGaaGOnaiaa % dAfaa8aabeaak8qacqGH9aqpcaWGsbWdamaaBaaaleaapeGaamiDai % aadIgaa8aabeaak8qacqGHxdaTdaWcaaWdaeaapeGaaG4naaWdaeaa % peGaaG4maaaacqGHRaWkcaaIXaGaey41aq7aaSaaa8aabaWdbiaaiE % daa8aabaWdbiaaiodaaaaaaa!4931! {V_{th,6V}} = {R_{th}} \times \frac{7}{3} + 1 \times \frac{7}{3}% MathType!End!2!1! \)
\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGH9aqpcaaIYaGaey41aq7aaSaaa8aabaWdbiaaiEdaa8aabaWd % biaaiodaaaGaey4kaSYaaSaaa8aabaWdbiaaiEdaa8aabaWdbiaaio % daaaGaeyypa0JaaG4naiaadAfaaaa!410C! = 2 \times \frac{7}{3} + \frac{7}{3} = 7V% MathType!End!2!1! \)
This is a linear network, so Vth at port B can be written as
Vth = V1 α + β
Where V1 is the input applied at port A.
We have V1 = 10 V, Vth,10 V = 9 V
9 = 10 α + β
when V1 = 6 V, Vth,6 V = 9 V
7 = 6 α + β
⇒ α = 0.5, β = 4
Thus, with any voltage V1 applied at port A, thevenin voltage or open circuit voltage at port B will be
\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGwbWdamaaBaaaleaapeGaamiDaiaadIgacaGGSaGaamOva8aa % daWgaaadbaWdbiaaigdaa8aabeaaaSqabaGcpeGaeyypa0JaaGimai % aac6cacaaI1aGaaiiOaiaadAfapaWaaSbaaSqaa8qacaaIXaaapaqa % baGcpeGaey4kaSIaaGinaaaa!43E8! {V_{th,{V_1}}} = 0.5\;{V_1} + 4% MathType!End!2!1! \)
V1 = 8 V
Vth,8 V = (0.5 × 8) + 4 = 8 = open circuit voltage
Important Questions from Two Port Networks
A two-port network is reciprocal if _________
A short-circuit admittance matrix of a two-port network is
\(\left[ {\begin{array}{} 0\\ {\frac{1}{2}} \end{array}\begin{array}{} { - \frac{1}{2}}\\ 0 \end{array}} \right]\)
The two-port network is
A two-port network has scattering parameters given \(\left[ s \right] = \left[ {\begin{array}{*{20}{c}} {{s_{11}}}&{{s_{12}}}\\ {{s_{211}}}&{{s_{22}}} \end{array}} \right]\). If the port 2 of the two-port is short-circuited, the s11 parameter for the resultant one-port network is
With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B is
In a linear two – port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short-circuited. When 5 V is applied to Port, a current of 1.25 A flows through a 1 Ω resistance connected across Port 2. When 3 V is applied to Port 1, then current (in Ampere) through a 2 Ω resistance connected across Port 2 is __________.
