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Current Electricity Applied Physics and Chemistry Circuits Lecture 1.
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Transcript of Current Electricity Applied Physics and Chemistry Circuits Lecture 1.
Current Electricity
Applied Physics and Chemistry Circuits Lecture 1
Electric Current
Current is flow of charges There must be a complete loop (closed) for flow
to occur Flow from high potential to low potential
As potential decreases, work is done by charges Conventional current is flow of positive charges Do positive charges actually flow?
Electric Current
Current may be direct or alternating Direct current: flows in one direction Alternating current: flows back and forth
Electric Circuit
Required for current electricity Closed loop or conducting path from high to low
potential Must have four parts:
Source of charges (area of high potential) Path for charges Device that reduces potential energy Sink (area of low potential)
Electric Circuit Energy is conserved in the circuit Charge is conserved in the circuit Energy carried by current depends on charge
and potential difference ΔE = qV Remember, V is the potential difference
1 V = 1 J/C
Electric Current
Current is rate of flow of electric charge Measured in Coulombs / sec 1 C/s = 1 A (Ampere) Symbol for current is I
Electric Power
Rate of energy transfer Measured in Watts 1W = 1 J/s
Electric Power
Since P = E/t And E = qV And q= It Then P = VI Power is current times the potential difference.
Equation for Amber
• In a circle!
Electric Power
If the current through a motor is 3 A and the potential difference is 120 V, what is the power of the motor?
Known: I=3A V=120 V Equation: P=IV P=(3A)(120V)=360 W
Electric Power Problem 2A 6 V battery delivers 0.5 A of current to an electric
motor. What is the power rating of this motor?
What we know:
V = 6V I = 0.5 A
Equation:
P = IV
Substitute:
P = (0.5 A)(6 V)
Solve!
P = 3 W
Electric Power Problem 2 Contd.How much energy does the motor use in 5.0
minutes?
What we know:
P = 3 W t = 5.0 min = 300 s
Equation:
P = W/t which means P = E/t and E = Pt
Substitute:
E = (3 W)(300 s)
Solve!
E = 300 J