1. QUIZ 4 PROBLEM 1 30V R 1, V 1 =6V R3R3 R2R2 1.5KΩ 2.7KΩ FIND I T, R T, V 3, V 2, R 3, P T, P 1, P 2, P 3 :SINCE IN A SERIES CIRCUIT

2. USING KIRCHOFF VOLTAGE LAW WE CAN CONFIRM VOLTAGE DROP CALCULATIONS 30V RtRt 7.5KΩ PROBLEM 1 CONTINUED QUIZ 4

3. PROBLEM 2 30 VR1R1 R2R2 R3R3 120Ω300Ω200Ω FIND I T, I 1, I 2, I 3, P T, P 1,P 2, P 3, GIVEN: FOR RESISTORS IN PARALLEL QUIZ 4

4. FIND(EQUILVANT OF R 1,R 2,R 3 ) =R T 30V RTRT 60Ω PROBLEM 2 CONTINUED QUIZ 4

5. PROBLEM 3 75V R1R1 R2R2 R6R6 R4R4 R5R5 R3R3 120Ω 560Ω 180Ω120Ω 68Ω FIND:R T, I T, V 1 TO V 6 I 1 TO I 6 FIRST FIND R T QUIZ 4 68Ω 560ΩR2R2 R3R3 STEP 2:REDRAW CIRCUIT

6. STEP 3:REDRAW CIRCUIT STEP 3 R EQ 140ΩR2R2 560Ω R1R1 120Ω R6R6 68Ω 75V PROBLEM 3 CONTINUED QUIZ 4

7. STEP 4:REDRAW 3TH TIME NOW WE CAN SOLVE FOR IN A SERIES CIRCUIT I IS THE SAME IN EACH RESISTOR USE KIRCHOFF’S VOLTAGE LAW TO CONFIRM THESE RESULTS 75V R1R1 R6R6 120Ω 68Ω R eq 2 112Ω 75V RTRT 300Ω PROBLEM 3 CONTINUED QUIZ 4

8. GOING BACK TO STEP 3 SOLVE FOR V2 FROM KIRCHOFF’S VOLTAGE LAW WE CAN LOOK AT THIS CIRCUIT AS HAVING TWO VOLTAGE LOOPS LOOP 1: SINCE V T,V 1,V 6 ARE KNOW WE CAN SOLVE FOR V 2 75V R1R1 R2R2 R6R6 R eq 2 120Ω 560Ω112Ω 68Ω LOOP1 LOOP2 V2V2 V6V6 PROBLEM 3 CONTINUED QUIZ 4 V1V1

9. GOING BACK TO STEP 2 USING KIRCHOFF’S CURRENT LAW, FIND I3, THEN FIND V3 KIRCHOFF’S CURRENT LAW STATES: THE SUN OF CURRENTS ENTERING A NODE (A) IS EQUAL TO ZERO.SO, I1I1 I3I3 I2I2 75V R1R1 R2R2 R6R6 R 4,5 120Ω 560Ω 72Ω 68Ω R3R3 PROBLEM 3 CONTINUEDQUIZ 4 R1R1 120 Ω 560Ω R2R2 R3R3 68Ω A A

10. ALL WE HAVE LEFT TO SOLVE FOR IS: FROM STEP 2, R3 AND R4,5 ARE IN SERIES,CURRENT IS THE SAME IN BOTH R3 AND R4,5 WORKING BACK TO THE ORIGINAL CIRCUIT GIVES, SINCE R4,R5 ARE IN PARALLEL, THE VOLTAGE IS THE SAME ACROSS EACH RESISTOR. 68Ω R3R3 R3 R 4,5 R4R4 R5R5 72Ω 68Ω 180Ω 120Ω B PROBLEM 3 CONTINUEDQUIZ 4

11. YOU CAN CONFIRM THIS USING KIRCHOFF’S CURRENT LAW FOR POINT B IN SUMMARY: R4R4 R5R5 B I5I5 I3I3 I4I4 PROBLEM 3 CONTINUED QUIZ 4