“Pumping” Charges: emf device Electromotive force (emf) device: does work on charges to bring them to higher potential to produce potential difference . Terminal voltage V ba Battery Tester R voltmeter >10MΩ r < 1 Ω

AC and House Wiring Generators/alternator: most influential emf device Amplitude: V m V rms =V m /sqrt(2)=120V V m =160V Power distribution system

Example Simple circuit A student built circuit as shown in the figure to power a electric circuit that requires 3V power input. The internal resistance of the circuit is R=50  The student used R 1 =100  and R 2 =300  He found the circuit did not function properly. How would you solve the problem by modifying the circuit R Without R: With R i2i2 i i1i1 For example: R 1 =5  R 2 =15 , V 1 =2.8V

 R2R2 Kirchhoff’s Voltage Law (KVL) The algebraic sum of potential difference in a loop must be zero 1.Draw current direction (arbitrary) and label the voltage direction (determined by the define current direction. 2. Define either clockwise or counterclockwise direction as positive voltage direction. Once the direction is defined, you have to use the same convention in every loop. (the sign for the voltage cross a resistor: + if current direction is the same as the loop direction, - other wise) I V1V1 V2V2 VrVr 3.Apply KVL: +V if V is in the same direction defined above. -  +V r +V 1 +V 2 =0  =V r +V 1 +V 2 =I(r+R 1 +R 2 ) I=  (r+R 1 +R 2 )

R2R2 R1R1 R3R3 Kirchhoff’s Voltage Law: multiloop The algebraic sum of potential difference in a loop must be zero 1.Draw current direction (arbitrary) and label the voltage direction (determined by the define current direction. 2.Define either clockwise or counterclockwise direction as positive voltage direction. Once the direction is defined, you have to use the same convention in every loop. I V1V1 V2V2 VrVr 3.Apply KVL: +V if V is in the same direction defined above. -  +V r +V 1 +V 2 =0 -V 2 +V 3 =0 V3V3 I2I2 I3I3  =Ir+IR 1 +I 2 R 2 -V 2 +V 3 =0

Kirchhoff’s Current Law (KCL) The algebraic sum of current at a node must be zero: I in =I out R2R2 R1R1 R3R3 I V1V1 V2V2 VrVr V3V3 I2I2 I3I3  =Ir+IR 1 +I 2 R 2 (2) V 3 -V 2 =0 (3) I=I 2 +I 3 (1)  3 V, r=1  R 1 =3  R 2 =5  R 3 =10  I - I 2 - I 3 = 0(4) 4I +5I 2 – 0I 3 = 3(5) 0I – 5I 2 +10I 3 = 0(6)

Last note on KVL & KCL If solutions to currents or voltages are negative, they mean the real directions are opposite to what you have defined!