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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

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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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