Question

4. (a)Find the Thevenin equivalent circuit for the network external to the resistor R in Figure 4. (b)Find the current through R when R is 22, 302, and 100 S2. Figure 4 5. (a)Find the Thevenin equivalent circuit for the network external to the resistor R in Figure 5. (b)Find the power delivered to R when R is 2 k2 and 100 3000 11 04ko 21.20 Figure 5 6. (a)Find the Thevenin equivalent circuit for the network external to the resistor R in Figure 6. (b)Find the power delivered to Rwhen Ris 2 2 and 100 Figure 6

Answer 1

SOLUTION 4 an M Mt o open the load erminal to find vines Since no current is on flowing through 7r. The cirenit can be modified is nov 12 r. 2 VTL ass voltage Lin so, by voltage divider: - r Th= Viar lov mi Wider: this resistance has -a no meaning in finding in H 1223 find Resistance 42 un na 122 phê Grin . Nih=7.sv]. 1. Rtha- y short circuit the voltage soure and across load - un is in parallel with 122 =) (4) 11/12) = 32 & 3л 18-ти - Rin=non The venin equivalent circuit. case a:- RL=24 I IL = 7.5 | 1220625A 7.Sve casebi. R2 =300 => IL = 76. (11= 0.1875 A1 case c': Re:100-2 IL = 7.5 = IL=0.06818 AL 102 M 12 ZR 10 +100

2. Applying source Transformation to the irrent source and 0.4kn resistor OLIKA 330-1 Er2k Vih Equivalent voltage = 3XO'UR IV = 12000) Now the circuit i become". - To find th, open the wad terminale 12000 € - No current flows through 330r resistor. so the circuit beomei 0.4 => using voltage divider 12001 Rule. =) Vin= Viak= (1.2k x 1200 1:24 TH 12K +0412 th=goov 3302 Tu • Rth - s same step as done in question t, short circuit the voltage source: . - 0.4k is in parallel with 1:2K E =) (0.414) 11(1.2k) - Boor our =) 3oon is in series with 330 i - Rth=6302 1 x Thevenin Equivalenti, Rth=6302 case a RL = 2 kr - . 630-42000 yoou G O I ZR2 = I2 = 0.342 2 A P= Il Ri =) 234.2089W) z

Casebi. RL =900kn IL= goo - 630+100K = 8.948103A IL=8.94m A =) PL = IERL » (p. guxio 3)² x 100k [Pe=&lalans 03) Thin I open the wad terminal and Saw in previous 2 Questions, here is of no use, as Iz2 = 0 so the circuit becomes; using voltage Divider =) V7-2= VTL I lov - Vih= ( 7 ) xov same as we 72 resistor Z72 Zan tt 7 [vith=sv= R.Th. - short circuit voltage source RAB = ? 773 T 72 ar ca RTL - Ar 1172) in series with 72 = 3.52 in series with 72

3.son M - Rth= 10.5 - 1 * The venin Equivalent: Rih=10 sen M With=svC case at Rizni. ZR2 SL 04A 10.5+2 IL = 5 ) 04A PL = 12 R? » (0.47²x22 IPL = 0.32 Watts case bi R2 = 100 ? IL = 5 0.04524 A '5 +100 PL = 12 R2 PL = 0.2047 watts -
The Thevenin theorem says that Any complicated linear network can be replaced by an equivalent voltage source (Vth) in series with Thevenin Resistance (Rth).

The aim here is to find the Thevenin voltage Vth and Thevenin Resistance Rth.

TO FIND Vth:

Open the load terminal and measure volatge across that terminal.

TO FIND Rth:

Short circuit all the Independent Voltage sources and Open circuit all the Independent current sources. Then find the equivalent resistance across the load terminal.

Here in the 2nd question we will use the concept of Source transformation in which a current source in paralllel with the resistance can be replaced by the voltage source in series with resistance where the value of Voltage source will be V=I*R.

I have tried my best to make you understand not only these questions but also the concept behind this.If you have any doubt feel free to ask in comments,I am always ready to help you.

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