1. Lecture Circuits: September 17, 2012 Page 1 Circuits Lab 01, Voltage and Resistors Thomas J. Crawford, PhD, PE July 12, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 1: Ohm’s Law : V = I * R Power: P = V * I = I 2 * R = V 2 / R Example: One resistor Given: V = 12.0 volts, R = 48.0 ohms (Ω ) Find: current I, power P. Solution: I = V / R = 12.0 / 48.0 = 0.25 amps = 250 milliamps P = V * I = 12.0 * 0.25 = 3.0 Watts Find: R if I = 0.0625 (1 / 16 ) amps. Solution: R = V / I = 12.0 / 0.0625 = 192 ohms ( Ω ) Example: Series Resistors Given: R1 = 50.0 ohms, R2 =200.0 ohms (Ω ) in series; V = 12.0 volts Find: current I, Δ V1, Δ V2, P1, P2. Solution: I = V / Req = 12.0 / (50. + 200.) = 0.048 amps = 48 milliamps Δ V1 = I * R1 = 0.048 * 50 = 2.4 volts, Δ V2 = I * R2 = 0.048 * 200 = 9.6 volts. Or by the Voltage Divider Rule Δ V1 = Δ V T * R1 / Req = 12.0 * 50.0 / 250.0 = 2.4 volts, Δ V2 = Δ V T * R2 / Req = 12.0 * 200.0 / 250.0 = 9.6 volts. P1 = I 2 * R1 = (0.048) 2 * 50.0 = 0.1152 watts P2 = I 2 * R2 = (0.048) 2 * 200.0 = 0.4608 watts Example: Parallel Resistors is on next page

2. Thomas J. Crawford, PhD, PE July 12, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 1: Ohm’s Law : V = I * R Power: P = V * I = I 2 * R = V 2 / R Example: Parallel Resistors Given: R1 = 21.0 ohms, R2 =42.0 ohms (Ω ) in parallel; V = 10.5 volts Find: Req, currents I, I1, I2 Solution: 1 / Req = 1 / R1 + 1 / R2 = 1 / 21 + 1 / 42  Req = 14.0 ohms I = I T = V / Req = 10.5 / 14.0 = 0.75 amps I1 = Δ V1 / R1 = 10.5 / 21 = 0.50 amps I2 = Δ V2 / R2 = 10.5 / 42 = 0.25 amps Or by the Current Divider Rule I1 = I T * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps I2 = I T * Req / R2= 0.75 * 14.0 / 42 = 0.25 amps Also next Page Lecture Circuits: September 17, 2012 Page 2 Circuits Lab 01, Voltage and Resistors

3. Thomas J. Crawford, PhD, PE July 12, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 1: Example: Parallel Resistors Given: R1 = 21.0 ohms, R2 =42.0 ohms (Ω ) in parallel; V = 10.5 volts Find: Req, currents I, I1, I2 Alternative Solution: Re-Writing the Req 1 / Req = 1 / R1 + 1 / R2  Req = R1 * R2 / ( R1 + R2 ) Req = 21.0 * 42.0 / ( 21.0 + 42.0 ) = 21.0 * 42.0 / 63.0  Req = 14.0 ohms Everything else remains the same: I = I T = V / Req = 10.5 / 14.0 = 0.75 amps I1 = Δ V1 / R1 = 10.5 / 21 = 0.50 amps and I2 = Δ V2 / R2 = 10.5 / 42 = 0.25 amps or Or I1 = I T * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps I2 = I T * Req / R2= 0.75 * 14.0 / 42 = 0.25 amps Note: This Alternative Solution gets complicated with 3 or 4 resistors: 3 resistors: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3  Req = R1 * R2 * R3 / ( R1* R2 + R1*R3 + R2*R3 ) 4 resistors: 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4  Req = R1 * R2 * R3 * R4 / ( you fill in the Rest !! ) Lecture Circuits: September 17, 2012 Page 3 Circuits Lab 01, Voltage and Resistors

4. Thomas J. Crawford, PhD, PE August 31, 2013 ENGR 1181_Lect_Circuits.ppt Illustrations: How are these resistors connected ? Illustrations: How are these resistors connected ? Illustrations: How are these resistors connected ? Illustrations: Relative to the discussion of Circuits, What do these symbols represent? Lecture Circuits: September 17, 2012 Page 4 Circuits Lab 01, Voltage and Resistors R1 R2 R3 R2 R1 R3

5. Thomas J. Crawford, PhD, PE August 31, 2013 ENGR 1181_Lect_Circuits.ppt Illustrations: How are these resistors connected ? Illustrations: How are these resistors connected ? A B R2 R3 R1 A B Lecture Circuits: September 17, 2012 Page 5 Circuits Lab 01, Voltage and Resistors R3 R2 R1

6. Thomas J. Crawford, PhD, PE August 31, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 1: Example: Given: The Circuit shown below; R3 = 80.0, R1 = ? R2 = ? ohms, V = 24.00 Volts Find: Current through R3 when the Switch is Open, and when the Switch is Closed. Solution: Apply Ohm’s Law, note the Parallel configuration. Switch Open: I3 = 0.00 amps Switch Closed: I3 = Δ V3 / R3 = 24.00 / 80.00  I3 = 0.300 amps Solution: We get the same result if we know the values of R1 & R2. Let R1 = R2 = 10.00 ohms  Req = 16.0  IT = 24 / 16 = 1.50  I3 = 0.30 amp Let R1 = R2 = 40.00 ohms  Req = 40.0  IT = 24 / 40 = 00.60  I3 = 0.30 a Let R1 = R2 = 360.00 ohms  Req = 72.0  IT = 24 / 72 = 1/3  I3 = 0.30 amp V = 24.0 R3 = 80R2 = ? Lecture Circuits: September 17, 2012 Page 6 Circuits Lab 01, Voltage and Resistors R1 = ?Switch

7. Thomas J. Crawford, PhD, PE August 31, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 1: Example: Given: The Circuit shown below; R1 = 80.0, R2 = 80.0 ohms, V = 24.00 Volts Find: Current through R1, R2 when the Switch is Open. Solution: Apply Ohm’s Law, note the position of the LED. I1 = I2 = 0.00 amps Find: Current through R1, R2 when the Switch is Closed. Solution: Apply Ohm’s Law, etc No current flows through R1, I1 0.00 amps I2 = V / R2 = 24.0 / 80.0  I2 = 0.30 amps V = 24.0 R2 = 80 R2 Lecture Circuits: September 17, 2012 Page 7 Circuits Lab 01, Voltage and Resistors R1 = 80 Switch

8. Thomas J. Crawford, PhD, PE July 26, 2013 ENGR 1182_Lect_RC_Circuits.ppt Examples: Given: R1 = 100 ohms, Δ V R1 = 2.80 volts Δ V LED = 2.45 volts, Switch is Open. Find: Voltage of the Power Supply or Battery. Solution: Ohm’s Law. V = Δ V R1 + Δ V LED = 2.80 + 2.45 = 5.25 volts R1R1 R1R1 R LED V V V R1R1 V Find: Current through R1. Solution: Ohm’s Law. I = Δ V R1 / R1 = 2.80 / 100. = 0.0280 amps Find: Resistance of the LED; R LED Solution: I is common to R1 and the LED; I = Δ V / R Hence Δ V R1 / R1 = Δ V LED / R LED  R LED = R1 * Δ V LED / Δ V R1 = 100 * 2.45 / 2.80  R LED = 87.50 ohms EXTRA DRAWING S Lecture Circuits: September 17, 2012 Page 8 Circuits Lab 01, Voltage and Resistors

9. Lecture Circuits: September 17, 2012 Page 9 Circuits Lab 02, Battery Resistance and Transmission Cables Thomas J. Crawford, PhD, PE February 16, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 2: Internal Resistance of Dry Cell Battery Example: Given: Open Circuit Voltage E = 1.624, Loaded Circuit Voltage V T = 1.584 External Resistor R = 79.2 ohms Find: Internal Resistance r, current I, and power P. Solution: r = R * ( E / V T – 1 ) = 79.2 * ( 1.624 / 1.584 – 1 ) = 2.00 ohms i = V T / R = 1.584 / 79.2 = 0.020 amps OR i = ( E – V T ) / r = ( 1.624 – 1.584 ) / 2.00 = 0.020 amps Various Powers Resistor: P = i 2 * R = (0.02) 2 * 79.2 = 0.03168 Watts P = V T 2 / R = (1.584) 2 / 79.2 = 0.03168 Watts Internal: P = i 2 * r = (0.02) 2 * 2.00 = 0.0008 Watts P = ( E – V T ) 2 / r = (1.624 – 1.584) 2 / 2.00 = 0.0008 Watts Total: P = i 2 * (R + r) = (0.02) 2 * ( 79.2 + 2.0) = 0.03248 Watts P = E 2 / (R + r) = (1.624) 2 / ( 79.2 + 2.0 ) = 0.03248 Watts

10. Lecture Circuits: September 17, 2012 Page 10 Circuits Lab 02, Battery Resistance and Transmission Cables Thomas J. Crawford, PhD, PE June 17, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 2: Transmission Cable: Example: Given: Single copper wire diameter d = 3.20 mm, L = 2400 m Find: Resistance of the single wire. Solution: R = ρ * L / A = 4 * ρ * L / ( π d 2 ) = 4 * 1.72E-08 * 2400 / π ( 0.0032 ) 2 = 5.133 ohms Example: Seven – strand Aluminum Transmission Cable Given: Wire diameter d = 2.80 mm, L = 1800 m Find: Resistance of the Transmission Cable. Solution: R = ρ * L / A = 4 * ρ * L / ( π d 2 ) Get Area first Area A = N * π * d 2 / 4 = 7 * π ( 2.8 ) 2 / 4 = 43.10 mm 2 R = 2.75E-08 * 1800 * 1.00E+06 / 43.10 = 1.148 ohms. Find: Max current the Cable may carry (Ampacity) if J = 2.20 amps / mm 2 Solution: i = J * A = 2.2 * 43.10 = 94.8 amps Given: The Transmission Cable Data Sheets Find or Show: the Percent Line Loss of I 2 * R = Percent Voltage Drop of Δ V Solution: Write out what each calculation is: Percent Line Loss = I 2 * R / Power = I 2 * R / ( V * I ) = I *R / V Percent Voltage Drop = Δ V / V = I * R / V, hence each percent is I * R / V Similarly, Percents for Copper and Aluminum are essentially equal because the Max Current Densities J are proportional to Resistivity ( ρ ).

11. Lecture Circuits: April 06, 2013 Page 11 Circuits Lab 03, DC Circuits with MATLAB. Thomas J. Crawford, PhD, PE July 12, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 3: Kirchhoff’s Rules and MATLAB: Example: Given: The Simple Circuit shown below; R1 = 80.0, R2 = 160.0 ohms. V = 12.00 V Find: Req, Current I, and illustrate Kirchhoff’s Laws or Rules. Solution: Apply Ohm’s Law and Kirchhoff’s Rules Node Rule: Summation of currents equals zero, or Currents in = currents out; Loop Rule: Summation of P.S. Voltages = summation of Voltage Drops. Req = R1 + R2 = 80.0 + 160.0  Req = 240.0 ohms I = V / Req = 12.0 / 240.0  I = 0.050 amps = 50. mAmps Kirchhoff’s Loop Rule: V – I1*R1 – I2*R2 = 0.0  12.0 – 0.05 * 80.0 – 0.05*160.0 = 0.0 ? Yes 12.0 – 4.0 – 8.0 = 0.0 Yes, it Does. V R1 = 80 R2 = 160 Δ V1 = I*R1 = 0.05 * 80 = 4.0 volts Δ V2 = I*R2 = 0.05 * 160 = 8.0 volts

12. Lecture Circuits: April 06, 2013 Page 12 Circuits Lab 03, DC Circuits with MATLAB Thomas J. Crawford, PhD, PE March 30, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 3: Five Resistors and Five Currents from P178. R1R1 R2R2 R5R5 R3R3 R4R4 V a b c d i1i1 i2i2 i3i3 i5i5 i4i4 I II III

13. Lecture Circuits: April 06, 2013 Page 13 Circuits Lab 03, DC Circuits with MATLAB. Thomas J. Crawford, PhD, PE May 13, 2013 ENGR1181_Lect_Circuits.ppt Circuits Lab 3: Kirchhoff’s Rules and MATLAB: Example: Given: The Circuit from P118, P-178 shown on the previous page. Find: The equations for the five currents in the circuit. Solution: Apply Kirchhoff’s Rules Node Rule: Summation of currents equals zero, or Currents in = currents out; Loop Rule: Summation of P.S. Voltages = summation of Voltage Drops. SEND THIS PAGE 13 TO STUDENTS. NEXT PAGE 14 HAS TOO MANY DETAILS AND RESULTS. MAY 13, 2013.