1. Kirchoff’s rules
And you

2. Rule #1 The Junction rule: The sum of currents coming to a junction =
the sum of the currents leaving a junction

3. Rule #2 The Loop Rule: The sum of all potential differences
around a loop = 0
That is to say the V battery = V drop total

4. Resistors in series Rtotal = R1 + R2… 6Ω 3Ω = 9Ω 6V 6V
6Ω Ω = Ω
6V V
(I) must be the same through both (= V/ΣR)
(V) calc. for each indiv. resistor = IR (for that resistor) R1 R2

5. The calculations Rtotal = 6Ω + 3Ω = 9Ω Itotal = Vtotal = 6 = .67 Amps
Vdrop1 = I1R1 = (.67)(6) = 4V
Vdrop2 = I2R2 = (.67)(3) = 2V *note sum to 6V !!!!!!!!

6. Basic Circuit 1 (like on test)

7. The answers R total = 10+5 = 15 ohms
I total = V/Rtotal = 10/15 = .67 Amps (same through each resistor)
V1 = (I1)(R1) = (.67)(10) = 6.7V
V2 = (I2)(R2) = (.67)(5) = 3.3V
*note these sum to 10V = Vtotal
I1 = I2 = Itotal = .67 Amps

8. Resistors in Parallel 1/Rtotal = 1/R1 + 1/R2… 6V 6V ? R total
4Ω = Ω 8Ω
6V V
? R total
? I branch 1 ?
? I branch 2 ?
? I total ?
Branch 1
Branch 2

9. 1/Rt = 1/4 + 1/8 = 3/8 = .375
so Rt = 1/ = Ω
I branch 1 = V/R1 = 6/4 = 1.5 Amps
I branch 2 = V/R2 = 6/8 = .75Amps
It = I1+I2 = 1.5Amps Amps
= Amps

10. Basic Circuit 2

11. The key 1/Rtotal = 1/R1 + 1/R2 = 1/10 + 1/5 = .3
So Rtotal = 1/.3 = ohms
Itotal = V/Rtotal = 10/ = 3 amps
V1 = V2 = Vtotal = 10 V

12. I1 = V1 / R1 = 10/10 = 1 amp
I2 = V2 / R2 = 10/5 = 2 amps
Note that the total current sums to 3 amps

13. ?Rtot, ? I1, ? I2, ? Itot,
Rtot : 1/Rtot = 1/R1+ 1/R2 = 1/ /5 = 3/10
So Rtot = 3.3Ω
I1 = V/R1 = 10/10 = 1 amp
I2 = V/R2 = 10/5 = 2 amps
So Itot = 3 amps
Or: Itot = V(1/Rtot) = (10)(1/3.3) = 3 amps!!!!

14. Basic Circuit 3

15. You try this one on your own and we will discuss it in a bit
Did you find:
Rtot = 3.6Ω Itot = 5.6 Amps
V1 = 10V V2 = 10V V3 = 20V
I1 = 3.3A I2 = 3.3A I3 = 2.2 A

16. Basic Circuit 4

17. Did you find Rtot = 5.25Ω Itot = 3.8Amps
I1 = 3.8 A I2 = 2.87 A I3 = .96 A
V1 = 11.4 V V2 = 8.6V V3 = 8.6V
Good Job!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!