1. p212c27: 1 Direct Current Circuits Two Basic Principles: Conservation of Charge Conservation of Energy Resistance Networks R eq I a b

2. p212c27: 2 Resistors in series Conservation of Charge I = I 1 = I 2 = I 3 Conservation of Energy V ab = V 1 + V 2 + V 3 a b I R 1 V 1 =I 1 R 1 R 2 V 2 =I 2 R 2 R 3 V 3 =I 3 R 3 Voltage Divider:

3. p212c27: 3 Resistors in parallel Conservation of Charge I = I 1 + I 2 + I 3 Conservation of Energy V ab = V 1 = V 2 = V 3 a I R 1 V 1 =I 1 R 1 R 2 V 2 =I 2 R 2 R 3 V 3 =I 3 R 3 b Current Divider:

4. p212c27: 4 R 1 =4  R 2 =3  R 3 =6  E =18V Example: Determine the equivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery

5. p212c27: 5 Kirchoff’s Rules Some circuits cannot be represented in terms of series and parallel combinations E1E1 E2E2 R1R1 R2R2 R3R3 R1R1 R3R3 R2R2 R4R4 R5R5 E Kirchoff’s rules are based upon conservation of energy and conservation of charge.

6. p212c27: 6 I1I1 I2I2 I3I3 Conservation of Charge: Junction rule The algebraic sum of the currents into any junction is zero.  I = 0 = + I 1  I 2  I 3  I = 0 =  I 1  I 2  I 3 R1R1 R2R2 R3R3 Alternatively:The sum of the currents into a junction is equal to the sum of the currents out of that junction. I 1  I 2 =  I 3

7. p212c27: 7 Conservation of Energy: Loop rule The algebraic sum of the potential differences around any closed loop is zero. I1I1 I2I2 I3I3 E1E1 E2E2 R1R1 R2R2 R3R3

8. p212c27: 8 Potential Differences in the direction of travel E +  direction of travel V =  E E +  direction of travel V = + E I +  direction of travel V =  IR I +  direction of travel V = + IR

9. p212c27: 9 I1I1 I2I2 I3I3 E1E1 E2E2 R2R2 R1R1 R3R3  I 1  I 2  I 3 =0  E 1 + I 1 R 1  I 2 R 2 + E 2 =0  I 3 R 3  I 2 R 2 + E 2 =0  I 3 R 3  I 1 R 1  E 1 =0

10. p212c27: 10 I1I1 I2I2 I3I3 E 1 =12V E 2 =5V R 2 =1  R 1 =2  R 3 =3   I 1  I 2  I 3 =0  12  I 1 2   I 2 1 + 5 =0  I 3 3  I 2 1+ 5 =0

11. p212c27: 11 Standard linear form:  I 1  1 I 2  1 I 3 = 0  2 I 1  1 I 2  0 I 3 = 7  0 I 1  1 I 2  3 I 3 = 5  I 1  I 2  I 3 =0  12  I 1 2   I 2 1 + 5 =0  I 3 3  I 2 1+ 5 =0 TI-89: [2nd] [MATH] [4](selects matrix) [5] (selects simult) Entry line: simult([1,1,-1;2,-1,0;0,1,3],[0;7;5]) I 1 = 3 I 2 =  1 I 3 = 2

12. p212c27: 12 R 1 R 2 R 3 I3A (-)1A 2A V P P E 1 P E 2 I1I1 I2I2 I3I3 E 1 =12V E 2 =5V R 2 =1  R 1 =2  R 3 =3 

13. p212c27: 13 Electrical Instruments Galvanometer: Torque proportional to current I Restoring torque proportional to angle => (Equilibrium) angle proportional to angle  Maximum Deflection =“Full Scale Deflection” I fs = Current to produce Full Scale Deflection R c = Resistance (of coil) V fs = I fs R c example: I fs = 1.0 mA, R c = 20.0  V fs = 20 mV =.020 V G

14. p212c27: 14 Wheatstone Bridge Circuit Balanced: V ab = 0 R2R2 b R1R1 R3R3 R4R4 E a G

15. p212c27: 15 Ammeter Measures current through device Measure currents larger than I fs by bypassing the galvanometer with another resistor (shunt Resistor R s ). I a = ammeter full scale When I = I a, we need I c = I fs V G = V sh => I fs R c = (I a -I fs )R sh G R sh I IcIc I-I c Design a 50 mA ammeter from example galvanometer I fs = 1.0 mA, R c = 20.0  V fs = 20 mV =.020 V

16. p212c27: 16 Voltmeter Measures potential difference across device Measure V larger than V fs by adding a resistor (R s ) in series to the galvanometer. V m = Voltmeter full scale When V = V m, we need I c = I fs I c = I s, V m = V s + V G => V m = I fs R s + I fs R c R s = (V m / I fs )- R c G RsRs I V Design a 10 V voltmeter from example galvanometer I fs = 1.0 mA, R c = 20.0  V fs = 20 mV =.020 V

17. p212c27: 17 Ohmmeter Measures resistance across terminals Meter supplies an EMF, resistance is determined by response. When R = 0, we need I c = I fs E = (I fs R s + I fs R c ) => V m = I fs R s + I fs R c R s = ( E / I fs )- R c G RsRs I E R

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20. p212c27: 20 Resistance Capacitance Circuits Charging Capacitor i E R C switch +q -q v c = q/C v R = iR Capacitor is initially uncharged. Switch is closed at t=0 Apply Loop rule: E -q/C-iR=0

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22. p212c27: 22 i E R C switch +q -q v c = q/C v R = iR

23. p212c27: 23 Resistance Capacitance Circuits Discharging Capacitor i R C switch +q -q v c = q/C v R = iR Capacitor has an initial charge Q o. Switch is closed at t=0 Apply Loop rule:  q/C-iR=0

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25. p212c27: 25 i R C switch +q -q v c = q/C v R = iR