DC Circuits: Ch 32

• Voltage – Starts out at highest point at “+” end of battery

• Voltage drops across lightbulbs and other sources of resistance.

• Voltage increases again at battery.

+ -

I

Voltage highest Voltage zero

The following circuit uses a 1.5 V battery and has a 15 W lightbulb.

a. Calculate the current in the circuit (P = IV)

b.Calculate the voltage drop across the lightbulb.

c. Sketch a graph of voltage vs. path (battery, top wire, resistor, bottom wire)

Resistors in Series

• Same current (I) passes through all resistors (bulbs)

• All bulbs are equally bright (energy loss, not current loss)

• Voltage drop across each resistor (V1,V2, V3)

V = V1 + V2 + V3

V = IR1 + IR2 + IR3

V = I(R1 + R2 + R3)

Req = R1 + R2 + R3

V = IReq

Resistors in Parallel

• Current splits at the junction

• Same Voltage across all resistors

I = I1 + I2 + I3

I1 = V

R1

I = V

Req

1 = 1 + 1 + 1

Req R1 R2 R3

Which combination of auto headlights will produce the brightest bulbs? Assume all bulbs have a resistance of R.

For the Bulbs in Series:

Req = R + R = 2R

For the Bulbs in Parallel

1 = 1 + 1

Req R R

1 = 2

Req R

Req = R/2

The bulbs in parallel have less resistance and will be brighter

What current flows through each resistor in the following circuit? (R = 100 )

Req = R1 + R2

Req = 200

V = IReq

I = V/Req

I = 24.0 V/ 200 = 0.120 A

Calculate the current through this circuit, and the voltage drop across each resistor.

Req = 400 + 290

Req = 690

V = IR

I = V/Req

I = 12.0 V/690 I = 0.0174 A

Vab = (0.0174A)(400)

Vab = 6.96 V

Vbc = (0.0174A)(290)

Vbc = 5.04 V

What current flows through each of the resistors in this circuit? (R = 100 )

1 = 1 + 1

Req 100 100

1/Req = 2/100

Req = 50

I = V/Req = 24.0 V/50 = 0.48 A

DC Circuits: Ex 4

What current will flow through the circuit shown?

1 = 1 + 1

Rp = 500 700

Rp = 290

Req = 400 + 290

Req = 690

V = IR

I = V/R

I = 12.0 V/690 I = 0.017 A or 17 mA

Example 4

Calculate the equivalent resistance in the following circuit.

DC Circuits: Ex 5

What current is flowing through just the 500 resistor?

First we find the voltage drop across the first resistor:

V = IR = (0.017 A)(400 )

V = 6.8 V

The voltage through the resistors in parallel will be:

12.0 V – 6.8 V = 5.2 V

To find the current across the 500 resistor:

V = IR

I = V/R

I = 5.2 V/500 = 0.010 A = 10 mA

DC Circuits: Ex 6

Which bulb will be the brightest in this arrangement (most current)?

Bulb C (current gets split running through A and B)

What happens when the switch is opened?– C and B will have the same brightness (I is constant

in a series circuit)

DC Circuits: Ex 5

What resistance would be present between points A and B?

(ANS: 41/15 R)

EMF and Terminal Voltage

• Batteries - source of emf (Electromotive Force), E (battery rating)

• All batteries have some internal resistance r

Vab = E – Ir

Vab = terminal(useful)voltage

E = battery rating

r = internal resistance

EMF: Example 1

A 12-V battery has an internal resistance of 0.1 . If 10 Amps flow from the battery, what is the terminal voltage?

Vab = E – Ir

Vab = 12 V – (10 A)(0.10 )

Vab = 11 V

EMF: Example 2

Calculate the current in the following circuit.

1/Req = 1/8 + ¼

Req = 2.7

Req = 6

Req = 8.7

1/Req = 1/10

Req = 4.8

Everything is now in series

Req = 4.8

Req = 10.3

V = IR

I = V/R

I = 9.0 V/10.3 I = 0.87 A

EMF: Example 2a

Now calculate the terminal(useful)voltage.

V = E – Ir

V = 9.0 V – (0.87 A)(0.50 )

V = 8.6 V

Grounded

• Wire is run to the ground

• Houses have a ground wire at main circuit box

• Does not affect circuit behavior normally

• Provides path for electricity to flow in emergency

Kirchoff’s Rules

1. Junction Rule - The sum of the currents entering a junction must equal the sum of currents leaving

2. Loop Rule - The sum of the changes in potential around any closed path = 0

Kirchoff Conventions

The “loop current” is not a current.Just a direction that you follow around the loop.

Kirchoff Conventions

Kirchoff’s Rule Ex 1

Junction Rule

I1 = I2 + I3

Loop Rule

Main Loop

6V – (I1)(4) – (I3)(9) = 0

Side Loop

(-I2)(5) + (I3)(9) = 0

I1 = I2 + I3 Eqn 3

6V – (I1)(4) – (I3)(9) = 0 Eqn 2

(-I2)(5) + (I3)(9) = 0 Eqn 3

Solve Eqn 1

(-I2)(5) + (I3)(9) = 0

(I3)(9) = (I2)(5)

I3 = 5I2

Substitution into Eqn 2

6V – (I2 + I3)(4) – (I3)(9) = 0

6 – 4I2 -4I3 - 9I3 = 0

6 – 4I2 - 13I3 = 0

I3 = 5I2 (from last slide)

6 – 4I2 - 13(5I2) = 0

6 = 101/9 I2

I2 = 0.53 A

I3 = 5/9 I2 = 0.29 A

I1 = I2 + I3 = 0.53 A + 0.29 A = 0.82 A

Kirchoff’s Rule Ex 2

I1 = I2 + I3

Loop Rule

Main Loop

9V – (I3)(10) – (I1)(5) = 0

Side Loop

(-I2)(5) + (I3)(10) = 0

(-I2)(5) + (I3)(10) = 0

(I2)(5) = (I3)(10)

I2 = 2I3

9V – (I3)(10) – (I1)(5) = 0

9V – (I3)(10) – (I2 + I3)(5) = 0

9 –10I3 – 5I2 – 5I3 = 0

9 –15I3 – 5I2 = 0

9 –15I3 – 5(2I3) = 0

9 –25I3 = 0

I3 = 9/25 = 0.36 A

I3 = 9/25 = 0.36 A

I2 = 2I3 = 2(0.36 A) = 0.72 A

I1 = I2 + I3 = 0.36A + 072 A = 1.08 A

Kirchoff’s Rules: Ex 3

Calculate the currents in the following circuit.

I1 + I2 = I3

Bottom Loop (clockwise)

10V – (6)I1 – (2)I3 = 0

Top Loop (clockwise)

-14V +(6)I1 – 10 V -(4)I2 = 0

Work with Bottom Loop

10V – (6)I1 – (2)I3 = 0

I1 + I2 = I3

10 – 6I1 – 2(I1 + I2) = 0

10 – 6I1 – 2I1 - 2I2 = 0

10 - 8I1 - 2I2 = 0

10 = 8I1 + 2I2

5 = 4I1 + I2

I2 = 5 - 4I1

Working with Top Loop

-14V +(6)I1 – 10 V -(4)I2 = 0

24 = 6I1 - 4I2

12 = 3I1 - 2I2

12 = 3I1 - 2(5 - 4I1)

22 = 11I1

I1 = 22/11 = 2.0 Amps

I2 = 5 - 4I1

I2 = 5 – 4(2) = -3.0 Amps

I1 + I2 = I3

I3 = -1.0 A

Batteries in Series

• If + to -, voltages add (top drawing)

• If + to +, voltages subtract (middle drawing = 8V, used to charge the 12V battery as in a car engine)

Batteries in Parallel

• Provide large current when needed

Extra Kirchoff Problems

I1 = -0.864 A

I2 = 2.6 A

I3 = 1.73 A

A Strange Example

Calculate I (-0.5 Amps (we picked wrong direction))

a. Calculate the equivalent resistance. (2.26 )

b. Calculate the current in the upper and lower wires. (3.98 A)

c. Calculate I1, I2, and I3 (0.60 A, 0.225 A, 1.13 A)

d. Sketch a graph showing the voltage through the circuit starting at the battery.

RC Circuits

• Capacitors store energy (flash in a camera)

• Resistors control how fast that energy is released

• Car lights that dim after you shut them

• Camera flashes

Vc + Vr = 0

Q - IR = 0 (Divide by R)

C

Q - I = 0 (I = -dQ/dt)

RC

Q + dQ = 0

RC dt

= time constant (time to reach 63% of full voltage)

= RC

A versatile relationship

V = Vo e-t/RC

I = Io e-t/RC

Q = Qo e-t/RC

I generally find voltage, then use V=IR and Q=VC

RC Circuits: Ex 1

What is the time constant for an RC circuit of resistance 200 k and capacitance of 3.0 F?

= (200,000 )(3.0 X 10-6 F) = 0.60 s

(lower resistance will cause the capacitor to charge more quickly)

RC Circuits: Ex 2

What will happen to the bulb (resistor) in the circuit below when the switch is closed (like a car door)?

Answer: Bulb will glow brightly initially, then dim as capacitor nears full charge.

RC Circuits: Ex 3

An uncharged RC circuit has a 12 V battery, a 5.0 F capacitor and a 800 k resistor. Calculate the time constant.

= RC

= (5.0 X 10-6 F)(800,000 W)

= 4.0 s

What is the maximum charge on the capacitor?

Q = CV

Q = (5.0 X 10-6 F)(12 V)

Q = 6 X 10-5 F or 60 F

What is the voltage and charge on the capacitor after 1 time constant?

V = (0.632)(12 V) = 7.584 V

Q = (0.632)(60 F) = 38 F

Consider the circuit below. Calculate:

a. The time constant (6 ms)

b. Maximum charge on the capacitor (3.6 C)

c. Time to reach 99% of maximum charge (28 ms)

d. Current when charge = ½ Qmax (300 A)

e. Maximum current (600 A)

f. The charge when the current is 20% of the maximum value. (2.9 C)

Discharging the RC Circuit

V = Vo e-t/RC

RC Circuits: Ex 4

An RC circuit has a charged capacitor C = 35 F and a resistance of 120. How much time will elapse until the voltage falls to 10 percent of its original (maximum) value?

V = Vo e-t/RC

0.10Vo =Voe-t/RC

0.10 =e-t/RC

ln(0.10) = ln(e-t/RC)

-2.3 = -t/RC

2.3 = t/RC

t = 2.3RC

t = (2.3)(120)(35 X 10-6 F)

t = 0.0097 s or 9.7 ms

RC Circuits: Ex 5

If a capacitor is discharged in an RC circuit, how many time constants will it take the voltage to drop to ¼ its maximum value?

V = Vo e-t/RC 1.39 = t/RC

0.25Vo =Voe-t/RC t = 1.39RC

0.25 =e-t/RC

ln(0.25) = ln(e-t/RC)

-1.39 = -t/RC

A fully charged 1.02 mF capacitor is in a circuit with a 20.0 V battery and a resistor. When discharged, the current is observed to decrease to 50% of it’s initial value in 40 s.

a. Calculate the charge on the capacitor at t=0 (20.4 C)

b.Calculate the resistance R (57 )

c. Calculate the charge at t = 60 s (7.3 C)

The capacitor in the drawing has been fully charged. The switch is quickly moved to position b (camera flash).

a. Calculate the initial charge on the capacitor. (9 C)

b.Calculate the charge on the capacitor after 5.0 s. (5.5 C)

c. Calculate the voltage after 5.0 s (5.5 V

d.Calculate the current through the resistor after 5.0 s (0.55 A)

Meters

• Galvanometer– Can only handle a small current

• Full-scale Current Sensitivity (Im)

– Maximum deflection

• Ex:– Multimeter– Car speedometer

Measuring I and V

Measuring Current– Anmeter is placed in series– Current is constant in series

Measuring Voltage– Voltmeter placed in parallel– Voltage constant in parallel circuits– Measuring voltage drop across a resistor

Anmeter (Series) Voltmeter (parallel)

DC Anmeter

• Uses “Shunt” (parallel) resistor

• Shunt resistor has low resistance

• Most of current flows through shunt, only a little through Galvanometer

• IRR = IGr

Meters: Ex 1What size shunt resistor should be used if a galvanometer has a full-scale sensitivity of 50 A and a resistance of r= 30 ? You want the meter to read a 1.0 A current.

Voltage same through both (V=IR)IRR = IgrSince most of the current goes through the shuntIR ~ 1 A

IRR = Igr

(1 A)(R) = (50 X 10-6 A)(30 )

R = 1.5 X 10-3 or 1.5 m

Meters: Ex 2

Design an anmeter that can test a 12 A vacuum cleaner if the galvanometer has an internal resistance of 50 W and a full scale deflection of 1 mA.

IRR = Igr

(12 A)(R) = (1 X 10-3 A)(50 )

R = 4.2 X 10-3 or 4.2 m

DC Voltmeter

• Resistor in series

• Large R for resistor (keeps current low in Galvanometer)

• V = I(R + r)

Meters: Ex 3What resistor should be used in a voltmeter that

can read a maximum of 15 V? The galvanometer has an internal resistance of 30 and a full scale deflection of 50 A.

V = I(R + r)

12 V = (50 X 10-6 A)(R + 30)

R + 30= 12 V

50 X 10-6 A

R + 30= 300,000 R ~ 300,000

Meters: Ex 4

Design a voltmeter for a 120 V appliance with and internal galvanometer resistance of 50 and a current sensitivity of 1 mA.

(ANS: R = 120,000 )

Electric Power

• Watt

• 1 Watt = 1 Joule

1 second

P = I2R

“Twinkle, twinkle, little star. Power equals I2R”

Power: Ex 1

Calculate the resistance of a 40-W auto headlight that operates at 12 V.

P = I2R

V = IR (so I =V/R)

P = V2R

R2

P = V2

R

P = V2

R

R = V2 = (12 V)2 = 3.6 P 40 W

Household Electricity

• Kilowatt-hour

• You do not pay for power, you pay for energy

1 kWh = (1000 J)(3600 s) = 3.60 X 106 J

1s

Power: Ex 2

An electric heater draws 15.0 A on a 120-V line. How much power does it use?

P = I2R

V = IR so R = V/I

P = I2V

I

P = IV = (15 A)(120V) = 1800 W or 1.8 kW

Power: Ex 3

How much does it cost to run it for 30 days if it operates 3.0 h per day and the electric company charges 10.5 cents per kWh?

Hours = 30 days X 3.0 h/day = 90 h

Cost = (1.80 kW)(90 h)($0.105/kWh) = $17

Power: Ex 4

A lightening bolt can transfer 109 J of energy at a potential difference of 5 X 107V over 0.20 s. What is the charge transferred?

V = PE/Q

Q = E/V = 109 J/ 5 X 107V = 20 C

What is the current?

I = Q/t

I = 20 C/0.2 s = 100 A

What is the power?

P = I2R

V = IR so R = V/I

P = I2V

I

P = IV = (100 A)(5 X 107V ) = 5 X 109 W

Household Electricity

• Circuit breakers – prevent “overloading” (too much current per wire)

• Metal melts or bimetallic strip expands

Household Electricity: Ex 1

Determine the total current drawn by all of the appliances shown.

P = IV

I = P/V

Ilight = 100W/120 V = 0.8 A

Ilight = 100W/120 V = 0.8 A

Iheater = 1800W/120 V = 15 A

Istereo = 350W/120 V = 2.9 A

Ihair = 1200W/120 V = 10.0 A

Itotal = 0.8A + 15.0A + 2.9A + 10.0A = 28.7 A

This would blow the 20 A fuse

DC vs AC

DC

•Electrons flow constantly

•Electrons flow in only one direction

•Batteries

AC

•Electrons flow in short burst

•Electrons switch directions (60 times a second)

•House current

DC vs AC

http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html

Jump Starting a Car

POSITIVE TO POSITIVE

(or your battery could explode)