Selected Student Solutions for Chapter 2

Transcription

1 /3/003 Assessment Prolems Selected Student Solutons for Chapter. Frst note that we know the current through all elements n the crcut except the 6 kw resstor (the current n the three elements to the left of the 6 kw resstor s ; the current n the three elements to the rght of the 6 kw resstor s 30 ). To fnd the current n the 6 kw resstor, wrte a KCL equaton at the top node: k 3 We can then use Ohm s law to fnd the oltages across each resstor n terms of. The results are shown n the fgure elow: a) To fnd, wrte a KVL equaton around the left-hand loop, summng oltages n a clockwse drecton startng elow the V source: - V + 4,000 -V + 86,000 0 Solng for, 4, ,000 6V 40, ,000 6V ma ) Now that we hae the alue of, we can calculate the oltage for each component except the dependent source. Then we can wrte a KVL equaton for the rght-hand loop to fnd the oltage of the dependent source. Sum the oltages n the clockwse drecton, startng to the left of the dependent source: Thus, -V , V - 86,000 40,000-8V 6V - 8V -V 40,000( 0-6 ) - 8V 0

2 /3/003 We now know the alues of oltage and current for eery crcut element. Let s construct a power tale: Element Current (ma) Voltage (V) equaton (mw) V p - - 4kW.3 p R 33.7 V p - - 6kW p R Dep. Source 77 - p kW 77.3 p R 0. 8V 77 8 p c) The total power generated n the crcut s the sum of the negate power alues n the power tale: -m W + -mw mW -60mW Thus, the total power generated n the crcut s 60 m W. d) The total power asored n the crcut s the sum of the poste power alues n the power tale: 33.7m W mW + 00mW + 0.mW 60mW Thus, the total power asored n the crcut s 60 m W..6 Gen that f A, we know the current n the dependent source s f 4A. We can wrte a KCL equaton at the left node to fnd the current n the 0W resstor. Summng the currents leang the node, - A + A + 4A + 0 0W A - A - 4A 0W -A Thus, the current n the 0W resstor s A, flowng rght to left, as seen n the crcut elow. a) To fnd s, wrte a KVL equaton, summng the oltages counter-clockwse around the lower rght loop. Start elow the oltage source. + (A)(0W) + (A)(30W) 0 - s - s + 0V + 60V 0 s 70V

3 /3/003 ) The current n the oltage source can e found y wrtng a KCL equaton at the rght-hand node. Sum the currents leang the node - A + A A -A 3A The current n the oltage source s 3A, flowng top to ottom. The power assocated wth ths source s p ( 70V)(3A) 0W Thus, 0W are asored y the oltage source. c) The oltage drop across the ndependent current source can e found y wrtng a KVL equaton around the left loop n a clockwse drecton: - + (A)(30W) 0 A - A + 60V 0 A 60V The power assocated wth ths source s p - A (-60V)(A) -300W Ths source thus delers 300W of power to the crcut. d) The oltage across the controlled current source can e found y wrtng a KVL equaton around the upper rght loop n a clockwse drecton: + + (0W)(A) 0 4A + 0V 0 4A -0V The power assocated wth ths source s p 4 A (-0V)(4A) -40W Ths source thus delers 40W of power to the crcut. e) The total power dsspated y the resstors s gen y ( ) (30W) + ( ) (0W) () (30W) + () (0W) A 30 W 0W 30W Chapter Prolems.30 To fnd the power asored e each resstor, we fnd the current through each resstor and use the equaton p R to calculate power. We lael the oltages and currents n the crcut as shown elow.

4 /3/003 We egn wth the.w resstor. Snce we know ts oltage, we can fnd ts current usng Ohm s law: 90V. 4A.W Next, we can fnd 4 y wrtng a KVL equaton around the outer loop of the crcut. Start elow the oltage source and sum the oltages n a clockwse drecton: - 40V + 90V V - 90V 0V Next, fnd the current usng Ohm s law: 0V 0A W Now fnd the current y wrtng a KCL equaton at the rght-most node. Sum the currents leang ths node A - 4A 6A Then fnd the oltage usng Ohm s law: (W)(6A) 30V Now wrte a KVL equaton for the top loop to fnd the oltage 4. Sum the oltages n the clockwse drecton. 90V V - 90V - 30V 60V Fnd the current 4 usng Ohm s law: 60V 4 A 4W Now wrte a KVL equaton for the ottom loop to fnd the oltage 0. Sum the oltages n the clockwse drecton V + 30V 80V Fnally, fnd the current 0 usng Ohm s law: 80V 0 9A 0W The oltages and currents found aoe are summarzed n the crcut elow.

5 /3/003 a) We can fnd the power asored y each resstor usng ether p R or p /R. Just to e consstent, we wll use p R: p () (4) 900W p (6) () 80W p. 4 (4) (.) 360W (9) (0) 60W p (0) () 00W ) To fnd the power delered y the oltage source, we must fnd the current n the oltage source, g. We can fnd ths current y wrtng a KCL equaton at the top node. Summng the currents leang, - + A + 4A 0 g g p 0 A + 4A 9A Fnd the power assocated wth the oltage source usng p -. (Note that the current g flows nto the mnus termnal of the oltage source). p40 V -(40)(9) -460W Thus, the oltage source supples 460 W to the crcut. c) The total power dsspated s the sum of the power alues assocated wth all of the resstors. 900 W + 80W + 360W + 60W + 00W 460W p dss.33 Frst note that we know the current through all elements n the crcut except the 00 W resstor (the current n the three elements to the left of the 00 W resstor s ; the current n the three elements to the rght of the 00 W resstor s 49 ). To fnd the current n the 00 W resstor, wrte a KCL equaton at the top node: We can then use Ohm s law to fnd the oltages across each resstor n terms of. The results are shown n the fgure elow: a) Our approach s to fnd frst, and then use the known alue of to fnd y. To fnd, wrte a KVL equaton around the left-hand loop, summng oltages n a clockwse drecton startng elow the 7.V source: 7.V +, V + 0,000 0 Solng for, -, ,000 6, ,000 6.V 6.V 00mA

6 /3/003 Now that we hae the alue of, we can calculate the oltage for each component except the dependent source. Then we can wrte a KVL equaton for the rght-hand loop to fnd the oltage of the dependent source. Sum the oltages n the clockwse drecton, startng to the left of the dependent source: - 4,00 + 9V -0,000 0 Thus, y.v. - Y Y 9V - 34,00-6 9V - 34,00(00 0 ) 9V - 3.4V.V ) We now know the alues of oltage and current for eery crcut element. Let s construct a power tale: Element Current (ma) Voltage (V) equaton (mw) 7.V p kw 00. p R 0 0.7V p 70 00W 000 p R 000 Dep. Source p 7,9 00W p R,00 9V p - -44,00 The total power generated n the crcut s the sum of the negate power alues n the power tale: - 79m W + -44,00mW -44,80mW Thus, the total power generated n the crcut s 44,80 m W. The total power asored n the crcut s the sum of the poste power alues n the power tale: 0m W + 70mW + 000mW + 7,9mW +,00mW 44,80mW Thus, the total power asored n the crcut s 44,80, whch equals the total power generated.