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Complex Circuits

Complex Circuits. Presentation 2003 R. McDermott. A power supply with an internal resistance (lowers terminal voltage): Adds an additional resistor in series with the power supply Reduces the voltage at the terminals Terminal voltage will depend on the external circuit.

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Complex Circuits

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  1. Complex Circuits Presentation 2003 R. McDermott

  2. A power supply with an internal resistance (lowers terminal voltage): Adds an additional resistor in series with the power supply Reduces the voltage at the terminals Terminal voltage will depend on the external circuit Example: A 12-volt battery has an internal resistance of 5.0 ohms. What is the terminal voltage of the battery when it is connected in series to a 50 ohm resistor? A 25 ohm resistor? Complex DC Circuits

  3. Complex Circuit #1 • Example: A 12-volt battery has an internal resistance of 5.0 ohms. What is the terminal voltage of the battery when it is connected in series to a 50 ohm resistor? • Rc = 5.0  + 50  = 55  • Ic = (12 v)/(55 ) = .22 a Internal drop = 5.0  X .22 a = 1.1 v Terminal voltage = 12 v – 1.1 v = 10.9 v

  4. Complex Circuit #2 • Example: A 12-volt battery has an internal resistance of 5.0 ohms. What is the terminal voltage of the battery when it is connected in series to a 25 ohm resistor? • Rc = 5.0  + 25  = 30  • Ic = (12 v)/(30 ) = .40 a Internal drop = 5.0  X .40 a = 2.0 v Terminal voltage = 12 v – 2.0 v = 10 v

  5. Complex Circuit #3 • Multiple batteries in a single loop: • Voltages add if the batteries are positive to negative • Voltages subtract if the batteries are positive to positive.

  6. Circuit #3 • The batteries are connected positive to positive, so their voltages are subtracted (they act against one another). • Vnet = 18v – 12v = 6v • The current in the circuit is then: • Ic = Vc/Rc = 6v/6 = 1a

  7. Complex Circuit #4 • Compound series and parallel circuit: • Both sets of rules apply on various parts of the circuit • Ratios become very important as time savers Find the current and voltage drops for each resistor.

  8. Circuit #4 • The parallel combination has a total resistance of: 1/Rp = 1/12 + 1/4 = 1/3 Rp = 3 ohms • The parallel combination is in series with R1 so: Rc = 6  + 3  = 9  • The circuit current can now be found: Ic = Vc/Rc = 18v/9 = 2a

  9. Circuit #4 (cont) • Resistor 1 gets the full circuit current, so its voltage drop is: V1= I1R1= (2a)(6) = 12 v • The parallel combination gets the remaining voltage: VP= Vc - V1= 18v - 12v = 6v • Each resistor in parallel gets the full 6 volts, so currents are: I2= VP/R2= 6v/4 = 1.5a I3= VP/R3= 6v/12 = 0.5a

  10. Complex Circuit #5 • Multiple loops with multiple batteries: Kirchoff’s junction current rule: IIN = IOUT. Kirchoff’s loop voltage rule: VLOOP = 0

  11. Circuit #5 • Kirchoff’s junction current rule: IIN= IOUT • Let’s pick junction A and assume all currents are downward. Let the far left wire be I1, the middle I2, and the far right be I3. All current is exiting the junction, so zero current enters (our assumption): • I1+ I2+ I3= 0 • We now have one equation which will allow us to solve for one unknown

  12. Circuit #5 (cont.) • Kirchoff’s loop voltage rule: VLOOP= 0 • We need two more equations, so we choose two of the three loops in our circuit (we have one on the left, one on the right, and the large outside loop) • Let’s take the left and right loops and arbitrarily pick a clockwise direction to follow • If the positive side of a battery pushes current with our loop, we assign the voltage a positive value. If it works against our loop, we assign a negative value

  13. Sample #5 • In the left loop, the 12v battery works with us (+) and the 6v works against us (–) • In the right loop, both batteries work with us (+) • Current flowing through a resistor in the same direction as we are looping produces a negative voltage. Current flowing opposite to our loop produces a positive voltage • I1 flows through R1 opposite to our loop, so I1R1 is positive. I2 flows through R2 in the same direction as our loop, so I2 R2 is negative

  14. #5 (cont.) • In the left loop, then: • 12 + I1R1– I2 R2– 6 = 0 and, 12 + 10I1 – 5 I2– 6 = 0 • This last is our second equation. We now have: I1+ I2+ I3= 0 12 + 10I1– 5 I2– 6 = 0

  15. #5 • For the right loop: • 6 + I2 R2– I3R3+ 22 = 0 and, 6 + 5I2– 20I3+ 22 = 0 • This is our third equation. We now have: I1+ I2+ I3= 0 12 + 10I1– 5 I2– 6 = 0 6 + 5I2– 20I3+ 22 = 0

  16. This is it! • Solving: • I1=– 0.657a I2=– 0.114a I3= 0.771a • We were wrong about the direction of I1 and I2 (both up), but correct for I3 (down)

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