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23-1 Simple Circuits

23-1 Simple Circuits. Series Circuits. Brightness of lamps is depended upon the ______________ in a circuit When all current travels in all of the devices, said to be in Series B/c current travels through all, current is same through out I = V source R A + R B

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23-1 Simple Circuits

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  1. 23-1 Simple Circuits

  2. Series Circuits • Brightness of lamps is depended upon the ______________ in a circuit • When all current travels in all of the devices, said to be in Series • B/c current travels through all, current is same through out • I = V source RA + RB • Equivalent Resistance • Sum of all individual resistance

  3. Problems • A 10 ohm resistor, 15 ohm resistor, and a 5 ohm resistor are connected in series across a 90 V battery. • A) What is the equivalent resistance in the circuit? • B) What is the current in the circuit? • Known Unknown • R1 = 10 ΩR = ?? • R2 = 15 Ω I = ??? • R3 = 5 Ω

  4. A) What is the equivalent resistance in the circuit? • R = RA + RB + RC • R = 10 Ω + 15 Ω + 5 Ω • R = 30 Ω

  5. B) What is the current in the circuit? • I = V source RA + RB + RC I = 90 V 10 Ω + 15 Ω + 5 Ω I = 3 A

  6. Voltage Drop • Voltage Divider • A Series circuit used to produce a voltage source of desired magnitude from a higher voltage battery • Solve by rearranging the R = V/I • V = IR • R is the equivalent resistance and V is the potential drop

  7. Suppose you have a 9-V battery but need a 5-V potential source. I = V/R = V / RA +RB VB = IRB VB = IRB = V x RB RA +RB VB = VRB RA +RB

  8. Two Resistors, 47 Ω and 82 Ω are connected in series across a 45.0 V Battery • A) What is the current in the circuit? • B) What is the voltage drop in each resistor? • C) The 47 Ω resistor is replaced by a 39 Ω resistor. Will the current increase, decrease, or stay the same? • D) What will happen to the voltage drop across the 82 Ω resistor? • Known Unknown • V source = 45.0 V I = ?? • RA = 47 Ω VA = ?? • RB = 82 Ω VB = ??

  9. A) What is the current in the circuit? • Find equivalent resistance • R = RA + RB • I = V source / R = V source / RA +RB • I = 45.0 V 47 Ω +82 Ω I = 0.349 A

  10. Find the voltage drop of each resistor VA = IRA = (0.349)(47 Ω) = 16 V VB = IRB = (0.349)(82 Ω) = 29 V

  11. Calculate Current again using RA and determine new voltage drop I = V source / R = V source / RA +RB • I = 45.0 V 39 Ω +82 Ω = 0.372 VB = IRB = (0.372)(82Ω) = 31 V

  12. A 9.0 V battery and two resistors, 400 ohms and 500 ohms, are connected as a voltage divider. What is the voltage across the 500 ohm resistor?

  13. Parallel Circuits • A circuit in which there are several current paths • Total current is the sum of the currents through each path and the potential difference across each path is the same • Current through path depends on each resistor

  14. Parallel Circuits • When resistance is removed, current through other resistors does not change, only the total current through the circuit

  15. Parallel Circuits • Finding Equivalent resistance • I = IA + IB + IC … • So then…. • V = V + V+ V … RA RB RC 1/R = (1/RA) + (1/RB) + (1/RC) …

  16. Three resistors, 60.0 Ω, 30.0 Ω, and 20.0 Ω, are connected in parallel across a 90.0 V battery • A) Find the current through each branch of the circuit • B) Find the equivalent resistance of the circuit • C) Find the current through the battery

  17. 2) Find the equivalent resistance • 1/R = (1/RA) + (1/RB) + (1/RC) • 1/R = (1/60 Ω) + (1/30 Ω) + (1/20 Ω) • = 0.100Ω-1 • R = 10.0 Ω • 3) Find the Total Current • I = V/R • = 90.0 V / 10.0 Ω • = 9.00 A • So we want the total current equal to the individual currents added together

  18. Find the current through each branch • IA = V / RA • = 90.0 V / 60.0 Ω = 1.50 A • IB = V / RB • = 90.0 V / 30.0 Ω = 3.00 A • IC = V / RC • = 90.0 V / 20.0 Ω = 4.50 A

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