# P=V 2 /R

### Parallel Circuit

• 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.

### Electric Current

• 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

### Example problem

• 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

### Potential Difference

• 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

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

• R = V/I

• Resistance = Potential difference/Current

• SI unit – ohm Symbol-

(omega)

### Example Problem

• 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

### Ohm’s Law

• 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

Electric power is the rate of conversion of electrical energy

• Formula for Electric Power:

P = IV

Electric power = current X potential difference

### Electric Power

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

### Electric Power

• 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

### Electric Power

• 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

### Energy Usage Cost

• 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)

### Example Problem

• 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