Current Proportionality and Thevenin Equivalence

[Novark]

Homework Statement

For the linear circuit shown, given that the current I in the 0.9 K\Omega is
10mA when Vs = 100V:

(a) Predict I using the proportionality property for: Vs = 25V, -12V, and 145 V respectively.

(b) Use Thevenin’s theorem across terminals a and b. Find I in terms of Vs and hence calculate I for the various values of Vs in (a).

Homework Equations

V = IR
Thevenin's Theorem

The Attempt at a Solution

I really don't know where to begin with this one. I'm lost for part (a), and for part (b), while I understand Thevenin's theorem, I'm just not seeing how to apply it to branch a-b.

If anyone can help me get going with this question, I would be very grateful.
Thanks.

 

[vk6kro]

The Thevenin part needs you to take out the 900 ohm resistor and see how the rest of the circuit would behave without it.

So, you could work out the voltages at a and b and get the difference between them if the 900 ohm wasn't there.

This is the Thevenin voltage.

To get the equivalent resistance in series with the Thevenin voltage you have to get the resistance looking into a and b without the power supply present (but shorted). This puts the resistors on each side of the circuit in parallel with each other, but in series with the other pair of parallel resistors. Don't forget the resistor values are in K's.

Knowing the Thevenin voltage and resistance, you can now put loads between a and b and see what current would flow.

Try it for 100 volts and see if you get the answer given for I. Then try the other volatges asked for in the question.

 

[hadiffina]

I is directly proportional to Vs

therefore I=BVs where B is a constant
the initial values were I=10mA when Vs=100V
10m=B(100)
B=(10m)/100
B=0.1m

thus I=(0.1m)Vs

Vs=25 I=(0.1m)*25 = 2.5mA

Vs=-12 I=(0.1m)*(-12) =-1.2mA

Vs=145 I=(0.1m)*145 = 14.5mA

using Thevenin :

to find Rth, Short the Vs voltage source and open 0.9kohm resistor


Rth = [9k//1k] + [3k//2k] = 2.1kohm

to find Vth, place the Vs voltage source back while
0.9kohm resistor remains open


using voltage divider : voltage drop on 9kohm resistor
V1 = (9k/10k) * Vs =0.9Vs


voltage drop on 3kohm resistor
V1 = (3k/5k) * Vs =0.6Vs

when refer to bottom loop, the equation below is derived

-0.9Vs + Vth + 0.6Vs = 0
Vth = 0.3Vs



use the thevenin equivalent circuit to attach with 0.9kohm load :
Rth(2.1kohm) is in series with 0.9kohm resistor therefore the
equivalent resistor is 3kohm

I = 0.3Vs/3k
I = (0.1m)Vs which is consistent with proportionality property

just plug in the different values of Vs and voila youll get the answer