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Simple Circuits Problems. Resistors in Series R eq = R 1 + R 2 + R 3 … Equivalent resistance equals the total of individual resistances in series. I =

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Presentation on theme: "Simple Circuits Problems. Resistors in Series R eq = R 1 + R 2 + R 3 … Equivalent resistance equals the total of individual resistances in series. I ="— Presentation transcript:

1 Simple Circuits Problems

2 Resistors in Series R eq = R 1 + R 2 + R 3 … Equivalent resistance equals the total of individual resistances in series. I = ΔV/R eq ΔV = IR 1 and ΔV = IR 2 V T = V 1 + V 2 + V 3 … Resistors in Parallel 1/R eq = 1/R1 + 1/R 2 + 1/R 3 … Equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship. I T = I 1 + I 2 + I 3 … I = ΔV/R eq ΔV =IR eq

3 Series and Parallel Resistors SeriesParallel Currentsame as totaladd to find total Potential Differenceadd to find totalsame as total

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6 Each Resistor =10 Ω Req = Req = 30 Ω I = V/R I = 12V / 30 Ω I = 0.4A 12V V = IR V = (0.4A)(10 Ω) V = 4 volts V T = 4V + 4V + 4V = 12V I = V/R I = 4V / 10 Ω I = 0.4A P = IV P = (0.4A)(4V) P = 1.6 W

7 Resistor 1 = 10 Ω Resistor 2 = 20 Ω Resistor 3 = 30 Ω Req = Req = 60 Ω I = V/R I = 12V / 60 Ω I = 0.2A 12V V = IR 1, V = IR 2, V = IR 3 V = (0.2A)(10 Ω) = 2V V = (0.2A)(20 Ω) = 4V V = (0.2A)(30 Ω) = 6V V T = 2V + 4V + 6V = 12V I = V/R 1, I=V/R 2, I=V/R 3 I = (2V)/(10 Ω) = 0.2A I = (4V)/(20 Ω) = 0.2A I = (6V)/(30 Ω) = 0.2A

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9 Each Resistor =10 Ω 1/ Req = (1/10)+(1/10)+(1/10) = 3/10 Req = 3.33 Ω I = V/R I = 12V / 3.33 Ω I = 3.6A V = 12V I = V/R I = 12V / 10 Ω I = 1.2A I T = 1.2A + 1.2A + 1.2A = 3.6A P = IV P = (1.2A)(12V) P = 14.4W 12V

10 Resistor 1 = 10 Ω Resistor 2 = 20 Ω Resistor 3 = 30 Ω 1/Req = 1/10+1/20+1/30 = / 1 Req = 1 / = 5.45 Ω I = V/R I = 12V / 5.45 Ω I = 2.2A V = 12V I = V/R I = 12V / 10 Ω = 1.2A I = 12V / 20 Ω = 0.6A I = 12V / 30 Ω = 0.4A I T = 1.2A + 0.6A + 0.4A = 2.2A 12V


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