EE301 Thevenin Solved Problems Handout

Below are four "fundamental" Thevenin circuits, each of which represents a basic scenario you may face when solving for the open-circuit voltage (Vrn) and the Thevenin Equivalent Resistance (Rrn) . For each of these circuits, answers for the Thevenin Equivalent Voltage, Thevenin Equivalent Resistance, voltage across the load resistor, current through the load resistor, and power dissipated by the load resistor are given.

Scenario 1

8800 1.36k0

8mA 9900.

'\QO ))_,

\_-----4------11 b

4. l:>co.w Tk\ftWt,.. ~"'~~kJ. U.ttt I

SOJ\.

a

b

Answers

V /'H = 7.92 V

RTH = 2.35 k0.

111 = 2.55 mA I

( Z.3'1(12.f 15c)

Scenario 2

4.2kn 980 a

7V 2.2kn

b

Answers

~H = 2.41 V

RTH = 1.54 kQ

I R,, = 460 µA

V11 = 1.70 V I.

~1 =4.97 µW I.

2.1 K.Jl,

Scenario 3

4.1kn

13 mA 1.9kQ

{ 1{p_~\Sr_ L~ oJ l.U WwJw1>

Z. ~clit ~o( frr1 .

[TH : \JJ -:. \J't.3KJ\.. : I l ' fl.~ .Jl<. /l.:

a

4.3kQ

b

: z. L\. )o <~. ~;oo : ~v --

I, 9 Kll,.

Answers

vm = 10.J v RTH = 2.50 kD..

l u = 2.06 mA L

VR = 5.15 v I.

Pii =10.6 mW I

= 2.o& w..A--

Scenario 4

6800

22 v 1.2ko

680/l.

IT ~I. . °' /.21(1!, ~ 21\/ ± £5

b

--

a

5700

b

ll:: £> .Q-ot >~d

2- 2. ';.

l~S o

Answers

vm = 10.0 v R1H = 310 n 111 = 18.0 mA

I.

~/ = 4.41 v I.

Pii = 79.4 m W I.

lt.&\)AA --

Errt :::. \(h = I). · Rs10JL:: le . a~\/