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• A parallel circuit is one that has two or more paths for the electricity to flow – similar to a fork in a river

• In other words, the loads are parallel to each other.

• If the loads in this circuit were light bulbs and one blew out there is still current flowing to the others.

• Current is the rate at which electric charges move through a given area.

• SI unit is the Ampere or Amp.

• 1 A = 1 C/s

• I = ΔQ/t

• Current = charge / time

• The current in a light bulb is 0.835 A. How long does it take for a total charge of 1.67 C to pass a point in the wire?

ΔQ = 1.6 C I = 0.835 A t= ?

I = ΔQ/t t = ΔQ/I t= 1.6C/0.835A

t= 2.00s

• The potential difference between the positive and negative ends of batteries:

• All AA, AAA, C, D Cell Batteries = 1.5 V

• The only difference is how long they produce the 1.5 V.

• Car battery = 12 V

• Positive and Negative slots of an electrical outlet = 120 V

• Resistance- The opposition to the flow of current in a conductor

**• R = V/I**

**• Resistance = Potential difference/Current**

• SI unit – ohm Symbol-

(omega)

• The resistance of a steam iron is 19.0 Ω. What is the current in the iron when it is connected across a potential difference of 120V?

• R= 19.0 Ω V= 120V I= ?

• R=V/I

• I=V/R

• I=120V/19.0 Ω

• I= 6.32 A

• Relates electric current, voltage and resistance.

• • Current = Voltage/Resistance (I=V/R)

• Current (I)

• flow of e-

• I = Q/t

• I = V/R

• Ampere (Amps)

• A

• Resistance (R)

• opposition to the flow of e-

• R = V/I

• Ohms

• Ω

• Voltage (V)

• push of e-

• V = PE/Q

• V = IR

• Volts

• V

• **Electric power **is the rate of conversion of electrical energy

• Formula for Electric Power:

**P = IV**

Electric power = current X potential difference

Because **P= IV **and **V=IR **we can also say;

P= IV = I(IR) = I 2 R

**P = I 2 R**

Or, because **I = V/R, **we can also say:

P = IV = (V/R)V = V 2 /R

**P=V 2 /R**

• An electric space heater is connected across a 120 V outlet. The heater dissipates 1320 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater.

• P = V 2 /R R = V 2 /P

• R = 120 2 /1320

**• R = 10.9 Ω**

• Power companies measure energy not power, using the **kilowatt-**

**hour **as the unit

• **One kilowatt-hour = **the energy delivered in 1 hour at the constant rate of 1 kW.

• To convert between kWh and the SI unit of Joule:

**• 1 kWh = 3.6 X 10 6 J**

• Power (P)

• • meas. of the rate at which electricity does work.

• • P = V x I

• • Watts (W)

• Energy (E)

• •meas. By electric meter.

• • E = P x t

• • kilowatt-hour

• (kWh)

• Cost ($)

• •cost of electricity

• • Cost = energy x kwh

• ($.07)

• How much does it cost to operate a 100.0 W light bulb for 24 h if electrical energy costs $0.080 per kWh?

• P= 100W = 0.100 kW; t= 24 h

• Energy = Pt = 0.100 kW*24 h = 2.4 kWh

• Cost = 2.4 kWh*$0.080 = $0.19