Unit 1.6 Series and Parallel Circuits_edited

Unit 1.6 Series and Parallel Circuits_edited - Series and...

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1 Unit 1: Basic Circuit Theory Unit 1.6: Series and Parallel Circuits Series and Parallel Circuits f In this lecture, you will learn about: ± Series circuits • Voltage Sources in series • Resistors in series • Voltage dividers ± Parallel circuits • Current Sources in parallel • Resistors in parallel • Current dividers ± Equivalent circuits ± Measuring Voltage and Current The Single Loop Circuit Let’s start with two lights connected to our battery as shown. This is a single loop circuit. The Single Loop Circuit + - 12 V a b Light 2 Light 1 From KCL: I S = I 1 = I 2 I S I 2 I 1 Elements in a circuit which carry the same current are in series . The Series Circuit Example + - 12 V I S R 2 R 1 What is the power absorbed by each circuit element? First, we need voltage references for the resistors. + - V 1 + - V 2 Must obey the passive sign convention! Now, invoke KVL: V 1 + V 2 -12 ±=±0 V 1 + V 2 = 12 What we really need to know is the series current, I S . The Series Circuit Example + - 12 V I S R 2 R 1 What is the series current I S ? + - V 1 + - V 2 Invoke Ohm’s Law: V 1 = I S R 1 V 2 = I S R 2 Then substitute: I S R 1 + I S R 2 = 12 I S (R 1 + R 2 ) = 12 I S = 12/(R 1 + R 2 ) V 1 + V 2 = 12 From KVL:
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2 The Series Circuit Example + - 12 V I S R 2 R 1 + - V 1 + - V 2 For the lights: P 1 = I S 2 R 1 P 2 = I S 2 R 2 For the source: P S = -12 I S What is the power absorbed by each circuit element? I S = 12/(R 1 + R 2 ) Where: Why is the source power negative? The Series Circuit Example + - 12 V I S R 2 R 1 + - V 1 + - V 2 30 = 36/ R L R L = 1.20 Known: P 1 = P 2 = P L = 30 W R 1 = R 2 = R L P L = I S 2 R L I S = 12/(R L + R L ) If the lights are identical and each absorbs 30 watts, what is the resistance of each light? Then: P L = R L (6/R L ) 2 What is I S ? The Series Circuit Example f Recap: ± Established references for voltage and current. ± Used KVL. ± Used Ohm’s Law. ± Solved for I S . ± With I S you can find everything else: • Voltages • Powers + - 12 V I S R 2 R 1 + - V 1 + - V 2 A Single Node-Pair Example The Single Node-Pair Circuit Elements in a circuit which have the same voltage across their terminals are in parallel . Use KVL to find V 1 and V 2 -12 + V 1 = 0 V 1 = V 2 = 12.0 V Single Node-Pair Circuit f What is the power absorbed by each circuit element? f We note that V 1 , I 1 , V 2 and I 2 are assigned consistent with the Passive Sign Convention f Use KCL to sum the currents at Node 1 ± -Is + I 1 + I 2 = 0 or Is = I 1 + I 2 f Use Ohm’s Law to find I 1 and I 2 ± I 1 = V 1 /R 1 = 12/R 1 . Similarly, I 2 = 12/R 2 ± Is = 12/R 1 + 12/R 2 = 12(1/R 1 + 1/R 2 ) Node 1
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3 Single Node-Pair Example What is the power absorbed by each circuit element? For the lights, P 1 = (V 1 ) 2 /R 1 = 12 2 /R 1 = 144/R 1 P 2 = (V 2 ) 2 /R 2 = 12 2 /R 2 = 144/R 2 For the source, Ps = -12Is Where: Is = 12(1/R 1 + 1/R 2 ) Again, why is the source power negative? Parallel Circuit Example f Recap: ± KVL shows that all elements have the same voltage.
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This note was uploaded on 08/27/2009 for the course EE 16200 taught by Professor Williamneal during the Fall '08 term at University of Texas at Austin.

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Unit 1.6 Series and Parallel Circuits_edited - Series and...

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