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Basic Circuit Problem__

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**
a) Find V _{a} (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[kŁ[],

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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V_{a} = (20k)(10m)

= (20 x 10^{3})(10
x 10^{-3})

= 200 [V]

Answer a) : V_{a} = 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[kŁ[] and resistor R2 = 2 [kŁ[] 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 [kŁ[]
**Answer**: Req for R1 and R2 is 3 [kŁ[]

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__For part.2 Parallel Resistors__

We can see that resistor R3 = 1[kŁ[] and resistor R4 = 2 [kŁ[] 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 [kŁ[]

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**Answer**: Req for R3 and R4 is approximate 0.67 [kŁ[]

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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 [kŁ[], R2 =2 [kŁ[], R3 = 1 [kŁ[], R4 = 2 [kŁ[], R5 = 1 [kŁ[], R6 = 5 [kŁ[].

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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 [kŁ[]

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1 / R6,2 = 1/2 + 1/5

= 5/10 + 2/10

= 7/10

R6,2 = 10/7

= 1.4286 [kŁ[]

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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[kŁ[]

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With the information we have until now; R1,5 = 0.5[kŁ[], R6,2 = 1.4286[kŁ[], R3,4 = 3[kŁ[].

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 [kŁ[]

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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 [kŁ[]

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**Answer:** Req = 1.1739 [kŁ[]

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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 [kŁ[]
(Shown on

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__Figure #3.5__

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**Written
by Kevin Tang** (Nov 2010)

**Copyright 2010
- KEVKEVWORLD.NET**