2- Ohm’s law and Power. The total amount of charge that passes through a wire at any point per unit time is referred to as : 1) current 2) electric potential.

2- Ohm’s law and Power

• The total amount of charge that passes through a wire at any point per unit time is referred to as :

• 1) current

• 2) electric potential

• 3) voltage

• 4) wattage

• The direction of convention current is the direction that :

• 1) negative charges would flow

• 2) positive charges would flow

• 3) electron flow

How is eclectic power calculated?

• Electric power used by a device is equal to the current times voltage:

• P = I V ( SI unit W)

• Car outlets ( used to charge your cell phone) s may be rated at 20 A, so that the circuit can deliver a maximum power. A car batter has 12 V. What is the power?

• P=IV P=IV= (20 A) (12 V)=240 W

• electric power may be expressed as volt-amperes or even kilovolt-ampere

• 1 kA V=1 kW⋅

Electrons move slowly through wires

Potential difference drive electron through a wire

• Ohm’s law relates the applied potential difference to the current produced and the wires resistance.

Applied voltage(V) = current (I) x resistance(R)

V=I R

R = V / I

SI unit: Ohm ( Greek letter Omega) Ω

1 Ω = 1 V/ 1 A

Resistance

• Collisions between electrons and atoms in the wire cause a resistance to the electron’s motion.

• Analogy example --- Friction of resisting the motion of a box sliding across a floor

• Or dense crowd reducing your waling speed

Resistor –small device used in circuits to provide particular resistance to current ( can represent resistance of a wire or

device)

• A potential difference of 24 - V is applied to a device that has a resistance of 150 - Ω. How much current flows through the resistor?

A potential difference of 24 - V is applied to a device that has a resistance of 150 - Ω. How much current flows through the resistor?

• I = V/R

• = 24 V / 150 Ω

• = .16 A

• Power (P) is the rate at which energy (PE) is moved

• P = PE / t PE = q V

• = q V/ t I = q / t

• P = I V

Ohms law V=IR for electric power

Apply ohms law to V=IR

• P = power

• P = I V

• = I ( I R)

• = I2 R

Apply ohms law to V=IR

• P= power

• P = I V

• = ( V/R) V

• = V2 / R

3 valid power equations

Form Ohms law V=IR

P = IV

P= I2V

P=V2R

• A 12 V battery is connected to a device that has 570 Ω resistance. How much energy is dissipated in the resistance in 65 s ?

• Known:• Voltage = 12 V • Resistance (R) = 570 Ω• Δ t = 65 s • ΔPE = ?

• P= V2/ R• =(12 V)2/ 560 Ω• =.25 W

• ΔPE = P (Δ t)• = ( .25 W) (65 s) • = 16 J

• What's this symbol called Ω ?