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Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display.1
Chapter 3Voltage and
Current Laws
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these two networks are equivalentthere are three nodes and five branchesa path is a sequence of nodesa loop is a closed (circular) path
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KCL: The algebraic sum of the currents enteringany node is zero.
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i A + i B + (i C ) + (i D ) = 0
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Current IN is zero:i A + i B + (i C ) + (i D ) = 0
Current OUT is zero:(-i A )+ (-i B ) + i C + i D = 0
Current IN=OUT:i A+ i B = i C + i D
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Find the current through resistor R3 if it is known thatthe voltage source supplies a current of 3 A.
Answer: i =6 A
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KVL: The algebraic sum of the voltages aroundany closed path is zero.
v1 + (-v 2 )+ (v 3 ) = 0
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Sum of RISES is zero (clockwise from B):v1 +(- v 2 ) + v 3 = 0
Sum of DROPS is zero (clockwise from B):(-v1 ) + v 2 + (-v 3 ) = 0
Two paths, samevoltage (A to B) :
v1 = (-v 3 ) + v 2
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Find v R2 (the voltage across R 2) and the voltage v x.
Answer: v R2 = 32 V and v x= 6 V.
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Example: find the current i x and the voltage v x
Answer: v x= 12 V and i x =120 mA
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Solve for the voltage v x and and the current i x
Answer: v x=8 V and i x= 1 A
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All of the elements in a circuit that carry thesame current are said to be connected in
series.
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Elements in a circuit having a common voltageacross them are said to be connected in
parallel .
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Calculate the power absorbed by each circuit element.
Answer:
p120V = 960 W, p 30 = 1920 W
pdep = 1920 W, p15 = 960 W Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display. 13
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Find the voltage v and the currents i1 and i2.
Answer: v = 2 V, i 1 = 60 A, and i 2 = 30 A
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Determine the value of v and the power supplied by theindependent current source.
Answer: v = 14.4 V, power from current source is 345.6 mW
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Voltage sourcesconnected in seriescan be combinedinto an equivalentvoltage source:
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Current sources connected in parallel can becombined into an equivalent current source:
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Our circuit models are idealizations that canlead to apparent physical absurdities:
Vs in parallel (a) and I s in series (c) can lead toimpossible circuits
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Using KVL shows:
Req = R 1 + R 2 + + R N
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Find i and the power supplied by the 80 V source.
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Answer: i = 3 A and p = 240 W supplied
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Using KCL shows:
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Two resistors in parallel can becombined using the
product / sumshortcut.
Connecting resistors in parallel makesthe result smaller :
0.5 min (R1, R2 ) < R 1||R 2 < min (R1 ,R2 )
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Resistors in series share the voltage applied tothem.
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Find v x
Answer: v x(t) = 4 sin t V
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Resistors in parallel share the current throughthem.
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Find i3(t)
Answer: i 3(t) = 1.333 sin t V
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