Basic Circuit Problem
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a) Find Va (in V) for the circuit shown on Figure#1.
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Figure#1
Solution for a) :
Given i = 10 [mA] and R = 20[k?[],
notice this circuit is a single loop circuit, therefore i = 10 [mA] everywhere on the circuit (with the given direction),
we can directly apply V = R i ... Ohm's Law
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Va = (20k)(10m)
= (20 x 103)(10 x 10-3)
= 200 [V]
Answer a) : Va = 200 [V]
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Notice: Start from question b), I will use Multism (EWB) circuit design software to draw all the circuit diagrams.
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b) Part1. Find equivalent resistor Req for R1 and R2 (shown on Figure #2 left)
Part2. Find equivalent resistor Req for R3 and R4 (shown on Figure #2 right)
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Figure#2
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Solution for b):
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For part.1 Series Resistors
We can see that resistor R1 = 1[k?[] and resistor R2 = 2 [k?[] are connect in "series".
For "n" resistors connect in series, the
equivalent resistance are... Req = R1 + R2
+ R3 + ... + Rn =
Ri
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Req = R1 + R2
= 1 + 2
= 3 [k?[] Answer: Req for R1 and R2 is 3 [k?[]
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For part.2 Parallel Resistors
We can see that resistor R3 = 1[k?[] and resistor R4 = 2 [k?[] are connect in "parallel".
For "n" resistors connect in parallel, relationship between resistors and equivalent resistor is... 1 / Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
1 / Req = 1/ R3 + 1/ R4
= 1/1 + 1/2
= 3/2
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Req = 2 /3
= 0.67 [k?[]
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Answer: Req for R3 and R4 is approximate 0.67 [k?[]
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Now, shall we try a little harder one?
c) Find equivalent resistor Req (between point A and B) for R1, R2, R3, R4, R5 and R6 (shown on Figure #3)
Figure#3
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Solution for C):
Given R1 = 1 [k?[], R2 =2 [k?[], R3 = 1 [k?[], R4 = 2 [k?[], R5 = 1 [k?[], R6 = 5 [k?[].
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Step 1. We can combine all the same point on the circuit,
for example, look at the diagram below (Figure#3.1).
Figure#3.1
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On the left hands side, we can combine point A and point B,
because there are no other sources (e.g. resistor, voltage source, diodes, ...etc) in between.
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On the right hands side, we CANNOT combine point C and point D,
because there is a source in between point C and D, there we cannot combine the points.
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Apply this method, we can now rearrange the circuit as below (shown on Figure#3.2)
Figure#3.2
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Step 2 . Now, base on Figure#3.2 shown above.
We can see that R1 and R5, R6 and R2 are in parallel, therefore, we can find equivalent resistor for them.
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1 / R1,5 = 1/R1 + 1/R5
= 1/1 + 1/1
= 2
R1,5 = 1/2
= 0.5 [k?[]
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1 / R6,2 = 1/2 + 1/5
= 5/10 + 2/10
= 7/10
R6,2 = 10/7
= 1.4286 [k?[]
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Also, we can see R3 and R4 is series connection, we can combine them too.
R3,4 = R3 + R4
= 1 + 2
= 3[k?[]
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With the information we have until now; R1,5 = 0.5[k?[], R6,2 = 1.4286[k?[], R3,4 = 3[k?[].
We can re-draw the circuit as following (Shown on Figure#3.3)
Figure#3.3
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Step 3. Base on Figure#3.3 shown above.
We can continue combine R1,5 and R6,2 since they connect in series.
R1,5,6,2 = R1,5 + R6,2
= 0.5 + 1.4286
= 1.9286 [k?[]
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Shown on Figure#3.4.
Figure #3.4
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Last Step . Base on Figure#3.4 shown above.
We only left two resistors (R1,5,6,2 and R3,4) in parallel, combine them and find the final result.
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1 / Req = 1/ R1,5,6,2 + 1/R3,4
= 1/1.9286 + 1/3
= 0.5185 + 0.3333
= 0.8518
Req = 1/0.8518
= 1.1739 [k?[]
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Answer: Req = 1.1739 [k?[]
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Check Answer
You may wish to use circuit design software "NI Multisim" (http://www.ni.com/multisim/ ) to do the circuit simulation.
The simulation result shown Req for circuit in Figure #3 is same as our numerical answer =1.174 [k?[] (Shown on Figure#3.5)
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Figure #3.5
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Written by Kevin Tang (Nov 2010)
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