Lecture Circuits: September 17, 2012 Page 1 Circuits Lab 01, Voltage and Resistors Thomas J....

Lecture Circuits: September 17, 2012 Circuits Lab 01, Voltage and Resistors

Thomas J. Crawford, PhD, PEJuly 12, 2013

ENGR1181_Lect_Circuits.ppt

Circuits Lab 1:

Ohm’s Law : V = I * R Power: P = V * I = I2 * R = V2 / 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, ΔV1, ΔV2, P1, P2.

Solution: I = V / Req = 12.0 / (50. + 200.) = 0.048 amps = 48 milliamps

ΔV1 = I * R1 = 0.048 * 50 = 2.4 volts,

ΔV2 = I * R2 = 0.048 * 200 = 9.6 volts.

Or by the Voltage Divider Rule

ΔV1 = ΔVT * R1 / Req = 12.0 * 50.0 / 250.0 = 2.4 volts,

ΔV2 = ΔVT * R2 / Req = 12.0 * 200.0 / 250.0 = 9.6 volts.

P1 = I2 * R1 = (0.048)2 * 50.0 = 0.1152 watts

P2 = I2 * R2 = (0.048)2 * 200.0 = 0.4608 watts

Example: Parallel Resistors is on next page



Circuits Lab 1:

Ohm’s Law : V = I * R Power: P = V * I = I2 * R = V2 / 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 = IT = V / Req = 10.5 / 14.0 = 0.75 amps

I1 = ΔV1 / R1 = 10.5 / 21 = 0.50 amps

I2 = ΔV2 / R2 = 10.5 / 42 = 0.25 amps

Or by the Current Divider Rule

I1 = IT * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps

I2 = IT * Req / R2= 0.75 * 14.0 / 42 = 0.25 amps Also next Page




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 = IT = V / Req = 10.5 / 14.0 = 0.75 amps

I1 = ΔV1 / R1 = 10.5 / 21 = 0.50 amps and

I2 = ΔV2 / R2 = 10.5 / 42 = 0.25 amps or

Or

I1 = IT * Req / R1= 0.75 * 14.0 / 21 = 0.50 amps

I2 = IT * 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 !! )


Thomas J. Crawford, PhD, PEAugust 31, 2013

ENGR 1181_Lect_Circuits.ppt

Illustrations:

How are these resistors connected ?

Illustrations:


Illustrations:


Illustrations:

Relative to the discussion of Circuits, What do these symbols represent?


R1 R2 R3

R2

R1 R3


ENGR 1181_Lect_Circuits.ppt

Illustrations:


Illustrations:


AB

R2

R3

R1

AB

Lecture Circuits: September 17, 2012

Circuits Lab 01, Voltage and Resistors

R3

R2

R1



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 = ΔV3 / 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 = 80 R2 = ?


R1 = ? Switch



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


R1 = 80 Switch


ENGR 1182_Lect_RC_Circuits.ppt

Examples:

Given: R1 = 100 ohms, ΔVR1 = 2.80 volts ΔVLED = 2.45 volts, Switch is Open.

Find: Voltage of the Power Supply or Battery.

Solution: Ohm’s Law. V = ΔVR1 + ΔVLED = 2.80 + 2.45 = 5.25 volts

R1

R1RLED

V

RLED

V

V

R1 RLED

V

Find: Current through R1.

Solution: Ohm’s Law. I = ΔVR1 / R1 = 2.80 / 100. = 0.0280 amps

Find: Resistance of the LED; RLED

Solution: I is common to R1 and the LED; I = ΔV / R

Hence ΔVR1 / R1 = ΔVLED / RLED RLED = R1 * ΔVLED / ΔVR1 = 100 * 2.45 / 2.80

RLED = 87.50 ohms

EXTRA DRAWING

S

Lecture Circuits: September 17, 2012

Circuits Lab 01, Voltage and Resistors

Lecture Circuits: September 17, 2012 Circuits Lab 02, Battery Resistance and Transmission Cables

Thomas J. Crawford, PhD, PEFebruary 16, 2013


Circuits Lab 2:

Internal Resistance of Dry Cell Battery

Example:

Given: Open Circuit Voltage E = 1.624, Loaded Circuit Voltage VT = 1.584

External Resistor R = 79.2 ohms

Find: Internal Resistance r, current I, and power P.

Solution: r = R * ( E / VT – 1 ) = 79.2 * ( 1.624 / 1.584 – 1 ) = 2.00 ohms

i = VT / R = 1.584 / 79.2 = 0.020 amps OR

i = ( E – VT ) / r = ( 1.624 – 1.584 ) / 2.00 = 0.020 amps

Various Powers

Resistor: P = i2 * R = (0.02)2 * 79.2 = 0.03168 Watts

P = VT2 / R = (1.584)2 / 79.2 = 0.03168 Watts

Internal: P = i2 * r = (0.02)2 * 2.00 = 0.0008 Watts

P = ( E – VT )2 / r = (1.624 – 1.584)2 / 2.00 = 0.0008 Watts

Total: P = i2 * (R + r) = (0.02)2 * ( 79.2 + 2.0) = 0.03248 Watts

P = E2 / (R + r) = (1.624)2 / ( 79.2 + 2.0 ) = 0.03248 Watts

Lecture Circuits: September 17, 2012 Circuits Lab 02, Battery Resistance and Transmission Cables

Thomas J. Crawford, PhD, PEJune 17, 2013


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 d2 ) =

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 d2 ) Get Area first

Area A = N * π * d2 / 4 = 7 * π ( 2.8 )2 / 4 = 43.10 mm2

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 / mm2

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 I2 * R = Percent Voltage Drop of ΔV Solution: Write out what each calculation is:

Percent Line Loss = I2 * R / Power = I2 * R / ( V * I ) = I *R / V

Percent Voltage Drop = ΔV / 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 ).

Lecture Circuits: April 06, 2013 Circuits Lab 03, DC Circuits with MATLAB.



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

ΔV1 = I*R1 = 0.05 * 80 = 4.0 volts

ΔV2 = I*R2 = 0.05 * 160 = 8.0 volts

Lecture Circuits: April 06, 2013 Circuits Lab 03, DC Circuits with MATLAB

Thomas J. Crawford, PhD, PEMarch 30, 2013


Circuits Lab 3:

Five Resistors and Five Currents from P178.

R1

R2

R5

R3

R4

V

a

b

c

d

i1

i2

i3

i5

i4

I

II

III

Lecture Circuits: April 06, 2013 Circuits Lab 03, DC Circuits with MATLAB.

Thomas J. Crawford, PhD, PEMay 13, 2013


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.

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