2- ohm’s law and power. the total amount of charge that passes through a wire at any point per...
TRANSCRIPT
• 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⋅
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
• 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