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Chapter E6 Analyzing Circuits Chapter E3 problems due today.

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Presentation on theme: "Chapter E6 Analyzing Circuits Chapter E3 problems due today."— Presentation transcript:

1 Chapter E6 Analyzing Circuits Chapter E3 problems due today.

2 Series Circuits – only one path for the current  I=I 1 =I 2 =I 3 =…  V=V 1 +V 2 +V 3 +…  R=R 1 +R 2 +R 3 +… RR RRR I I I I I I I V V1V1 V2V2 V3V3 V4V4 As you go around the series circuit, add (or subtract) voltages. A B A B voltage voltages resistors

3 A more realistic view of voltage in a series circuit. Battery Thick wire Thin wire Back To battery Voltage Resistance

4 Series Circuits – only one path for the current R1R1 R2R2 R5R5 R4R4 R3R3 I I I IA Find the current in this circuit R 1 =1Ω R 2 =2Ω R 3 =3 Ω R 4 =4Ω R 5 =5Ω V=10v R = 15 Ω I = 0.666 A

5 Parallel circuits – current has a choice Battery R1R1 R2R2 R3R3 R4R4 I1I1 I2I2 I4I4 I6I6 I7I7 I3I3 I5I5 I1I1 I3I3 I5I5 I7I7 I7I7 I 1 =I 2 +I 3 I 3 =I 4 +I 5 I 5 =I 6 +I 7 R is the single resistor that will have the same resistance as the other 4 in parallel Battery R

6 Parallel circuits – Find the current through the battery. Battery R1R1 R2R2 R3R3 R4R4 I1I1 I2I2 I4I4 I6I6 I7I7 I3I3 I5I5 I1I1 I3I3 I5I5 I7I7 I7I7 V = 10 v R 1 =1Ω R 2 =2Ω R 3 =3 Ω R 4 =4Ω I=20.8 A I 5 = ? I 5 = 5.83 A

7 Battery R1R1 R2R2 R3R3 R4R4 I1I1 I2I2 I4I4 I6I6 I7I7 I3I3 I5I5 I1I1 I3I3 I5I5 I7I7 I7I7 V = 10 v R 1 =1Ω R 2 =2Ω R 3 =3 Ω R 4 =4Ω

8 Find the current through the battery and the 4 Ω resistor. The battery = 10v, R 1 = 1 Ω, etc. Battery R1R1 R2R2 R3R3 R4R4 I3I3 First find the equivalent resistance of the bottom 3 resistors. 1.55 Ω 0.61 ΩThe total resistance is I B =16.5 A I 4 =1.43 A

9 Electrical Safety  0.0005 amps – can feel  0.005 amps – painful  0.01 amps – causes muscles to contract  0.018 amps – stops breathing  Note that you were shocked by 300,000 volts in lab without harm.  Current kills, not voltage.  But higher voltage has the potential for higher current.

10 Problems due Wednesday  E6B.2, E6B.4, E6B.5, E6B.9


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