Presentation on theme: "Applied Physics and Chemistry Circuits Lecture 1"— Presentation transcript:
1 Applied Physics and Chemistry Circuits Lecture 1 Current ElectricityApplied Physics and Chemistry Circuits Lecture 1
2 Electric Current Current is flow of charges There must be a complete loop (closed) for flow to occurFlow from high potential to low potentialAs potential decreases, work is done by chargesConventional current is flow of positive chargesDo positive charges actually flow?
3 Electric Current Current may be direct or alternating Direct current: flows in one directionAlternating current: flows back and forth
4 Electric Circuit Required for current electricity Closed loop or conducting path from high to low potentialMust have four parts:Source of charges (area of high potential)Path for chargesDevice that reduces potential energySink (area of low potential)
5 Electric Circuit Energy is conserved in the circuit Charge is conserved in the circuitEnergy carried by current depends on charge and potential differenceΔE = qVRemember, V is the potential difference1 V = 1 J/C
6 Electric Current Current is rate of flow of electric charge Measured in Coulombs / sec1 C/s = 1 A (Ampere)Symbol for current is I
7 Electric PowerRate of energy transferMeasured in Watts1W = 1 J/s
8 Electric Power Since P = E/t And E = qV And q= It Then P = VI Power is current times the potential difference.
10 Electric PowerIf 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 VEquation: P=IVP=(3A)(120V)=360 W
11 Electric Power Problem 2 A 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
12 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 sEquation:P = W/t which means P = E/t and E = PtSubstitute:E = (3 W)(300 s)Solve!E = 300 J