1 / 15

Combination Circuits

Combination Circuits. Steps to Solve Combined Series-Parallel Circuits 1. If necessary, draw a diagram of the circuit. 2. Find any parallel resistors in the circuit and simplify them into one equivalent resistance using the formula for parallel equivalent resistance.

romaine-cesar
Télécharger la présentation

Combination Circuits

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Combination Circuits

  2. Steps to Solve Combined Series-Parallel Circuits 1. If necessary, draw a diagram of the circuit. 2. Find any parallel resistors in the circuit and simplify them into one equivalent resistance using the formula for parallel equivalent resistance. 3. If necessary, draw a new diagram using the equivalent resistor instead of the multiple previous resistors. 4. Find any resistors that are now in series and replace them with the equivalent resistance using the formula for series equivalent resistance. 5. If necessary, draw a new diagram using the equivalent resistance. 6. Once the circuit is reduced into a single resistor, you can now solve for the current using Ohm’s Law.

  3. Calculate the following: • total equivalent resistance • total current • the current across each resistor • the voltage drop across each resistor

  4. Draw the Circuit

  5. Solve for Req for parallel resistors 1/Req = 1/4 + 1/12 1/Req = .333 Req = 3 Ω Remember, the first step in combination circuits is ALWAYS to calculate the equivalent resistance of the parallel resistors!

  6. Redraw the Circuit 5 Ω 24 V 3 Ω 8 Ω

  7. Solve for Req for series resistors Req = 8 + 3 + 5 Req = 16 Ω Note: the 3Ω resistor came from the result of our solving for the Req for the parallel circuit section 5 Ω 24 V 3 Ω 8 Ω

  8. Redraw the Circuit 24 V 16 Ω

  9. Solve for the Total Current Vt = (It)(Rt) 24 = It(16) It = 1.5 amps Ohm’s Law: V = IR

  10. Solve for the Current through Each Resistor Since resistors R1 and R4 are in series, the current in series-connected resistors is the same everywhere. Therefore, It = I1 = I4 = 1.5 amps Note: In a Series Circuit, to solve for total current: It = I1 = I2 = I3 =…

  11. Solving for the Current through Each Resistor Since resistors R2 and R3 are in parallel, the current in parallel-connected resistors is added up to equal the total current. Therefore, It = I1 + I4 = 1.5 amps However, this gets a bit tricky because the resistors do not have the same value; therefore we must first calculate the voltage drop through each resistor and then come back to calculate the current

  12. Series Circuit, to solve for total voltage: Vt = V1 + V2 + V3 +… Calculate the voltage drop across the series-connected resistors. (R1 and R4 in diagram) V1 = I1R1 V4 = I4R4 V1 = (1.5)(5) = 7.5 V V4 = (1.5)(8) = 12 V

  13. Next, subtract the values for the series voltage from the total voltage VT – Vseries = Vparallel 24 V – 7.5 V – 12 V = 4.5 V This tells us that the voltage drop across EACH parallel resistor is 4.5 V because Vt = V1 = V2 = V3 = …

  14. Lastly, using Ohm’s Law calculate the current traveling through each parallel resistor V2 = I2R2 V3 = I3R3 4.5 = I2(4) 4.5 = I3(12) I2 = 1.125 amps I3 = .375 amps Remember, current varies through each parallel resistor since there is more than one path for the electrons to take!

  15. Results of our calculations:

More Related