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QUIZ 4
PROBLEM 1
FIND IT, RT, V3, V2, R3, PT, P1,P2, P3
R1 , V1 =6V
I1 
1.5KΩ
30V
FIND I T :SINCE IN A SERIES CIRCUIT
I T  I1  I 2  I 3
2.7KΩ
R2
V1
6V

 4mA  I T  I 2  I 3
R1 1.5K
FIND RT FROM OHM ' S LAW
RT 
VT 30V

 7.5K
I T 4mA
FIND R3
R3
RT  R1  R2  R3
R3  RT  R1  R2
R3  7.5K  1.5K  2.7 K
R3  3.3K
FIND V2 , V3
V2  I 2 R2  4mA  27k  10.8V
V3  I 3 R3  4mA  3.3k  13.2V
PROBLEM 1 CONTINUED
QUIZ 4
USING KIRCHOFF VOLTAGE LAW WE CAN CONFIRM VOLTAGE DROP CALCULATIONS
SINCE VT  V1  V2  V3
30V  6V  10.8V  13.2V
30V  30V
PT  I T VT  4mA  30V  120mW
P1  I1V1  4mA  6V  24mW
P2  I 2V2  4mA 10.8V  43.2mW
30V
P3  I 3V3  4mA 13.2V  52.8mW
AGAIN PT  P1  P2  P3
120mW  24mW  43.2mW  52.8mW
120mW  120mW
7.5KΩ
Rt
PROBLEM 2
FIND IT, I1, I2, I3, PT, P1,P2, P3, GIVEN:
QUIZ 4
30 V
120Ω
R1
200Ω
R2
300Ω
R3
FOR RESISTORS IN PARALLEL
VT  V1  V2  V3
SO FIND I1 , I 2 , I 3
V
30V
I1  1 
 250mA
R1 120
I2 
V2
30V

 150mA
R2 200
I3 
V3
30V

 100mA
R3 300
I T  I1  I 2  I 3
I T  250mA  150mA  100mA  500mA
PROBLEM 2 CONTINUED
QUIZ 4
FIND
R EQ
(EQUILVANT OF R1,R2,R3) =RT
ONE WAY : REQ 
ANOTHER WAY
VT
30V

 60
I T 500mA
REQ 
1
1

 60  RT
1
1
1
1
1
1




R1 R 2 R3 120 200 300
PT  I T VT  500mA  30V  15W
P1  I1V1  250mA  30V  7.5W
P2  I 2V2  150mA  30V  4.5W
30V
P3  I 3V3  100mA  30V  3W
NOTE : PT  P1  P2  P 3  7.5W  4.5W  3W  15W
60Ω
RT
PROBLEM 3
QUIZ 4
FIND:RT, IT, V1 TO V6 I1 TO I6
FIRST FIND RT
R3
68Ω
R4
R5
R1
120Ω
120Ω
R2
75V
180Ω
560Ω
STEP 2:REDRAW CIRCUIT
R6
68Ω
R4,5 
R4  R5
R4  R5

120 180 21,600

 72
120  180
300
2
R2
560Ω
R3
68Ω
R4,5
72
STEP 3:REDRAW CIRCUIT
SINCE R3  R4,5 ARE IN SERIES
PROBLEM 3 CONTINUED
REQ  R3  R4,5  68  72  140
R1
120Ω
R
75V
560Ω
REQ
140Ω
2
R6
68Ω
STEP 3
FIND REQ2 FOR R2  REQ
REQ 2 
R2  REQ
R2  REQ

560 140
 112
560  140
QUIZ 4
PROBLEM 3 CONTINUED
QUIZ 4
STEP 4:REDRAW 3TH TIME
NOW WE CAN SOLVE FOR
R1
RT  R1  REQ 2  R6
120Ω
RT  120  112  68
Req2 112Ω
75V
RT  300
NOW FIND I T FROM OHM ' S LAW
R6
IT 
68Ω
VT
75V

 0.25 A
RT 360
IN A SERIES CIRCUIT I IS THE SAME IN EACH RESISTOR
I R1  I REQ2  I R 6  I T
I1  0.25 A V1  I1 R1  .25 A 120  30V
I REQ2  0.25 A VREQ2  I REQ2 RREQ2  .25 A 112  28V
RT
75V
I 6  0.25 A V6  I 6 R6  .25 A  68  17V
USE KIRCHOFF’S VOLTAGE LAW TO CONFIRM THESE RESULTS
VT  V1  VREQ2  V6
VT  30V  2.8V  17V
VT  75V
300Ω
R1
PROBLEM 3 CONTINUED
V1
120Ω
GOING BACK TO STEP 3
SOLVE FOR V2
V2
LOOP1
R
75V
LOOP2
560Ω
2
QUIZ 4
R6
V6
68Ω
FROM KIRCHOFF’S VOLTAGE LAW WE CAN LOOK AT THIS CIRCUIT
AS HAVING TWO VOLTAGE LOOPS
LOOP 1:
VT  V1  V2  V6
SINCE VT,V1,V6 ARE KNOW WE CAN SOLVE FOR V2
V2  VT  V1  V6
V2  75V  30V  17V
V2  28V
FIND I 2
I2 
V2
28V

 0.05 A
R2 560
Req2
112Ω
QUIZ 4
PROBLEM 3 CONTINUED
R1
GOING BACK TO STEP 2 USING KIRCHOFF’S
CURRENT LAW, FIND I3, THEN FIND V3
R3 68Ω
A
120Ω
R4,5
R
75V
72Ω
560Ω
2
R6
KIRCHOFF’S CURRENT LAW STATES: THE SUN OF
CURRENTS ENTERING A NODE (A) IS EQUAL TO
ZERO.SO,
68Ω
I1  I 2  I 3
SINCE I1 , I 2 ARE KNOWN ,
SOLVE FOR I 3  I1  I 2
I 3  .25 A  0.05 A  0.20 A
SO V3  I 3 R3
V3  .2 A  68
V3  13.6V
R1
I1
I3
A
120
Ω
I2
R
2
560Ω
R3
68Ω
ALL WE HAVE LEFT TO SOLVE FOR IS:
I 4 , I 5 AND V4 ,V5
QUIZ 4
PROBLEM 3 CONTINUED
FROM STEP 2, R3 AND R4,5 ARE IN SERIES,CURRENT IS THE SAME IN BOTH R3 AND R4,5
SO I 3  I R 4,5  .2 A
VR 4,5  I R 4,5  R4,5
R3
VR 4,5  .2 A  72
68Ω
VR 4,5  14.4V
WORKING BACK TO THE
ORIGINAL CIRCUIT GIVES,
R3
R4,5
72Ω
120Ω
R4
68Ω
B
R5 180Ω
SINCE R4,R5 ARE IN PARALLEL, THE VOLTAGE IS THE SAME ACROSS EACH RESISTOR.
V4,5  VR 4  V5  14.4V
I4 
V4
14.4V

 0.12 A
R4
120
I5 
V5
14.4V

 0.08 A
R5
150
YOU CAN CONFIRM THIS USING KIRCHOFF’S CURRENT LAW FOR POINT B
I3  I 4  I5
QUIZ 4
I3
.2 A  .12 A  .08 A
I4
B
I5
.2 A  .2 A
PROBLEM 3 CONTINUED
IN SUMMARY:
RT  300
I T  0.25 A
V1  30V
I1  I T  0.25 A
V2  28V
I 2  0.05 A
V3  13.6V
I 3  .2 A
V4  14.4V
I 4  0.12 A
V5  14.4V
I 5  0.08 A
V6  17V
I 6  .25 A
R4
R5