With 10 V dc connected at port A in the linear non-reciprocal 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
C
Correct answer
Explanation
on applying Thevenin's equation, for (i) and (ii), we get
Vth = 3Rth +3.....(1)
Vth = 2Rth +5...(2)
subtracting (1) and (2), we get, Rth = 2ohms and Vth = 9V when input was 10V
if the input is 6V then Vth = 7/3 * (2+1) =7V
Vth = Vb=f(Vs) which implies Vth = mVs + c
Vth(10V) = m(10) +C...(3)
Vth(6) = m(6)+ C...(4)
Vth(7) = m(7) + C....(5) and Vth(8V) = m(8) + C.....(6)
by substituting values in eq. (3 & 4) we get,
9 = 10m +C...(7) and 7 = 6m + C...(8)
subtracting(7) and (8), we get, m = 0.5 and c = 4
now, Vth(8V) = m(8)+C = 8 volts