Solve for part A and B. I think I got part A correct, but I\'m not sure. Can som

Question

Solve for part A and B.

I think I got part A correct, but I'm not sure.

Can someone check if my ASSUMPTIONS ARE CORRECT?

From the circuit, I assumed:

ib = i1

40ib = -i2

Vo = 3000i2 (2)

For Part(A):

KVL@i1:

-Vs 1000i1 + 2000i2 + 3000(i1 - i2) = 0

i2 = (Vs/1700) - (13/17)i1 (1)

(1)(2) Vo = 3000[ (Vs/1700) - (13/17)i1]

Vo/Vs = (30 - 39000i1)/17

Is this correct for part A???.......I'm not sure because we also know that 40ib=-i2 and ib=i1 so I dont know if we should should for i2 or i1 for KVL@i1

Please help!

Thanks!

Explanation / Answer

assume that the current are in mA unit so that it will be easier to mulitply them with the kilo ohm units.,we get simple volt unit.

from the figure,it is clear that

ib=i1

and 40*ib=-i2

now in mesh 1,

vs=(2+1+.3)*i1-0.3*i2

putting i2=-40*ib=-40*i1

we get

vs=3.3*i1+12*i1=15.3*i1....(1)

vo=3*i2=-120*i1...(2)

so vo/vs=-120/15.3=-7.84

so gain of the amplifier 7.84

part B:

ib=i1

vs=15.3 i1

so input resistance:(put in mind that i1 is in mA where as vs is in volt)

input resistance=vs/ib(the given formula in the question is wrong...resistance=voltage/current NOT current/voltage)

=15.3/0.001=15.3 kilo ohm(the 0.001 comes due to fact that current is in mA unit)

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