Question

[p2bs 20q6a] The Thevenin or open circuit voltage "seen by" the load resistor RL is most nearly 2.4 KS2 1.6 ks2 L w 60 v +1 sa 24.8 k2 15 ma 25k2 : SRL W 1.8 ks2 6V None of the other answers 8V 10 V At

Answer 1

ow, voltage across skr is open circuit voltage seen by load. .: Using loop analysis 8 - for couprli 132 = 2.41; +fit (2) 418 or, 132=7.2eit 48 iz or, 11 = 0.6 ei toinez - For 2008-2 negative Lopposite direction - 72 = {1+6+5+48 (iztei) or, 72 = 4.811 + 11.4 ie 0 on, 36 = 2.42; + 5.7 iz (2) solving these two equations, i=-1195mA rendicates il = 19.63 MA so, current through Skd is in [email protected] mm directing ..open circuit Voltage = Voltage across SKR 2 1.95mA X 5 kr = 9.75 Volt so, most nearly to lov By

2-4 kr 1.6 , 115mA {sks & R2 First we convert "15mA current source with Parall 4-8 kr resistance" into Voltage soure for of Voltage, V= 15x16°x48X103 = 72 volt with series 4.8 kr resistance. 1882 2.4k2 1.6ks an 6ord. I 72 G skri & R2 i &4.8k2. i. 100p-1 1.8kr InoP-2

2.4.2 ---- . 1.682 2R 60V 4.8Wh Wisma 5423 - The current source of boxed portion is to be converted into voltage source. This voltage source has the voltage, V = IR = 15 mA x 4.8 K2 = 72 Volt 48k2 & 15mA * £ 720 4.8 6.12 Then circuit becomes as, 2.4K 1-6k2 Loop-2 600 t o 721 T 1 11, 418 KB Giz $5k2 i8 R₂ 100p-1 18k2 Y Now, as the load is opened, then open circuit voltage as seen by Ri will be voltage across 5 k2 resistance. Now, we can find this by using loop analysis. Using KVL in loop-l, 1932 = 2.48, + (litis) 4-8 or, .132= 7.2 i +4. 82 Or, 11=0.6 7 0.4 iz 0 For loop-2, 72= iz (1.6+5+18) + liitiz) 4.8 on, 72 = 4.8 i; + 13.2 iz or, 6 = 0.4i, +1.1 in +3) Solving equation (1) and (2) we get, is=-16 ma. 1. Open circuit voltage, Vith = 106 mA x 562 = 8 volt -(ams)

ow, voltage across skr is open circuit voltage seen by load. .: Using loop analysis 8 - for couprli 132 = 2.41; +fit (2) 418 or, 132=7.2eit 48 iz or, 11 = 0.6 ei toinez - For 2008-2 negative Lopposite direction - 72 = {1+6+5+48 (iztei) or, 72 = 4.811 + 11.4 ie 0 on, 36 = 2.42; + 5.7 iz (2) solving these two equations, i=-1195mA rendicates il = 19.63 MA so, current through Skd is in [email protected] mm directing ..open circuit Voltage = Voltage across SKR 2 1.95mA X 5 kr = 9.75 Volt so, most nearly to lov By