Help
Options
Didn't know it?
click below
click below
Knew it?
click below
click below
Don't Know
Remaining cards (0)
Know
retry
shuffle
restart
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
Normal Size Small Size show me how
Parallel Ciruits 5
Chapter 5
| Question | Answer |
|---|---|
| Branch | One current path in a parallel circuit |
| Current Divider | A parallel circuit in which the currents divide inversely proportional to the parallel branch resistances |
| Kirchhoff's Current Law | A law stating that the total current into a node equals the total current out of the node. Equivalently, it can be stated that the algebraic sum of all the currents entering and leaving a node is 0 |
| Node | A point or junction in a circuit at which two or more components are connected |
| Parallel | The relationship between two circuit components that exists when they are connected between the same pair of nodes |
| Total Parallel Resistance 5-1 | RT = 1÷ 1/R1 + 1/R2 + 1/R3 + ... 1/Rn |
| Special Case for Two Resistors in Parallel 5-2 | RT = R1R2 ÷ R1 + R2 |
| Special Case for n Equal-Value Resistors in Parallel 5-3 | RT = R ÷ n |
| Kirchhoff's Current Law 5-4 | Iin(1) + Iin(2) + Iin(3) + ... Iin(3) = Iout(1) + Iout(2) + Iout(3) + ... Iout(m) |
| General Current-Divider formula 5-5 | Ix = (RT ÷ Rx) IT |
| Two-Branch Current-Divider Formula 5-6 | I1 = (R2 ÷ R1 + R2) IT |
| Two_Branch Current-Divider Formula 5-7 | I2 = (R1 ÷ R1 + R2) IT |
| Total Power 5-8 | PT = P1 + P2 + P3 + ... Pn |
Created by:
phrogger
Popular Math sets