HWK34s_11G(1)

# HWK34s_11G(1) - V R 2 = IR 2 =(1 A(40Ω = 40 V 6 What is...

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Homework 34: Series Circuits: SOLUTIONS (Due Wednesday 3/28/07) SAR Physics AM Kolchin All work, equations, and units must be shown to get full credit. Below is a series circuit with three resistances. The resistances are: R 1 = 20 Ω R 2 = 40 Ω R 3 = 60 Ω A voltage of: V T = 120 V is maintained between points a and d. 1

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1. Calculate R EQ for this circuit by summing up the resis- tances. R EQ = R 1 + R 2 + R 3 = 20Ω + 40Ω + 60Ω = 120Ω 2. What is the current ( I ) through this circuit? I = V T R EQ = 120 V 120Ω = 1 A 3. What is the voltage drop V R 1 across R 1 ? (Apply Ohm’s Law between points a and b.) V R 1 = IR 1 = (1 A )(20Ω) = 20 V 4. What is the power used by R 1 ? P R 1 = V R 1 I = (20 V )(1 A ) = 20 W 5. What is the voltage drop V R 2 across R 2 ? (Apply Ohm’s Law between points b and c.)
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Unformatted text preview: V R 2 = IR 2 = (1 A )(40Ω) = 40 V 6. What is the power used by R 2 ? P R 2 = V R 2 I = (40 V )(1 A ) = 40 W 7. What is the voltage drop V R 3 across R 3 ? (Apply Ohm’s Law between points c and d.) V R 3 = IR 3 = (1 A )(60Ω) = 60 V 8. What is the power used by R 3 ? P R 3 = V R 3 I = (60 V )(1 A ) = 60 W 9. Add the voltage drops across R 1 and R 2 and R 3 . V R 1 + V R 2 + V R 3 = 20 V + 40 V + 60 V = 120 V 2 10. What is the total power drawn from the voltage source? P T = P R 1 + P R 2 + P R 3 = 20 W + 40 W + 60 W = 120 W . or P T = V T I = (120 V )(1 A ) = 120 W 3...
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HWK34s_11G(1) - V R 2 = IR 2 =(1 A(40Ω = 40 V 6 What is...

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