Question

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed.
(i) $1\Omega$ connected at port B draws a current of 3 A.
(ii) $2.5\Omega$ connected at port B draws a current of 2 A.


For the same network, with 6 V dc connected at port A, $1\Omega$ 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

• A. 6 V
• B. 7 V
• C. 8 V
• D. 9 V

Answer 1

The correct option is C 8 V


Case (i) :

$V_{dc}=10V,R_L=1\Omega ,I=3A$

$V_{Th}=I{R}_{Th}+R_L=3R_{Th}+1$

$\Rightarrow V_{Th}=3R_{Th}+3$ .....(i)

Case (ii) :

$R_L=2.5\Omega ,I=2A,V_{dc}=10V$

$V_{Th}=2(R_{Th}+2.5)$

$V_{Th}=2R_{Th}+5$ ......(ii)

From equations (i) and (ii)

$V_{Th}=9Vand{R}_{TH}=2\Omega$ ....(iii)

Now, $V_{Th}$ depends on independent voltage source and varies with applied voltage. $R_{Th}$ does not depend on voltage source and is same for any applied voltage source, since voltage source is short circuited while calculating $R_{Th}$.

For $V_{dc}=6V,R_L=1\Omega ,I=7/3A$

$V_{Th}=I{R}_{Th}+R_L$

$V_{Th}=\frac{7}{3}(2+1)=7V$ ...(iv)

$\because$The network is linear and non reciprocal, it may contain dependent voltage source.

$\therefore$ $V_{Th}=aV+b$ ..... (v)

$\Rightarrow$ 9 = a10 + b from equation (iii)

and 7 = a6 + b from equation (iv)

Solving, we get a = 1/2 and b = 4

$\therefore V_{Th}=\frac{V}{2}+4=\frac{8}{2}+4=8V$

$\therefore$ Open circuit voltage at port B = $8V$