Circuit Theory Assignment - Resistance & Power (Spring 2012), Exercises of Physics

The solution to assignment 1 of the circuit theory course (phy301) for the spring 2012 semester. The assignment required students to find the equivalent resistance of a given circuit and calculate the power dissipated through a specific resistance in the network. Detailed calculations and diagrams to help students understand the concepts.

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2011/2012

Uploaded on 08/03/2012

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Assignment 1(Spring 2012)
(Solution)
Circuit Theory (Phy301)
Marks: 20
Due Date: 13th April, 2012
Q 1:
Find the equivalent resistance of given circuit. Write each step of the calculation to get
maximum marks. Draw the circuit diagram of each step otherwise you will lose your
marks.
Sol:
Starting from right side we see that there is a short circuit (Direct wire without any
resistance) current will not pass through 12 resistance but follow easy path (short
circuit path), so ignoring the effect of this resistance circuit can be redrawn as
5 and 3 are in series
So R = 5 + 3 =8
Circuit adopts the shape as
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Assignment 1(Spring 2012)

(Solution)

Circuit Theory (Phy301) Marks: 20 Due Date: 13th^ April, 2012 Q 1: Find the equivalent resistance of given circuit. Write each step of the calculation to get maximum marks. Draw the circuit diagram of each step otherwise you will lose your marks.

Sol: Starting from right side we see that there is a short circuit (Direct wire without any resistance) current will not pass through 12Ω resistance but follow easy path (short circuit path), so ignoring the effect of this resistance circuit can be redrawn as

5  and 3 are in series So R = 5+ 3=8  Circuit adopts the shape as

4  and 4 are also in series So R = 4+ 4=8

At upper side 8 is parallel with 8 so 8||8 = (8 x 8)/(8+8) = 64/ = 4

6  is parallel with 6 so 6||6 = (6 x6)/6+6) = 36/ = 3

We see that 8A and 1.2Ω are in series so same current will pass through 1.2Ω resistance; however 8A current is divided to 4Ω and energy saver of 12Ω resistance. For understanding and simplification the above circuit can be redrawn as

Since we have to find power dissipation in 12Ω energy saver, that can be calculated by formula P=I 2 R So first we find current flowing through 12Ω resistance by current divider rule as

(^1) ( ) 1

1 4 2 12 8

4 8 2 16

(^2 ) 2 12 48

R I x I R T Here R R I A Putting above

x A

So power P

I R watt

    

 

 

Q 3: Answer the following questions. I. Why Copper wire is frequently used for conducting current? Ans: Atomic No. of Copper atom is 29, only 1 electron is available in outer N shell, when many such copper atom come close to each other, these unstable outer shell electrons are easily migrated from one atom to other randomly, called free electrons So due to easily availability of free electrons in copper, it is more preferred for conductivity. More free electrons cause more/easy conductivity. Moreover it bend easily, is ductile (is easily bent repeatedly without breaking). As the silver is not economical so the copper is widely and frequently used for conducting current.

II. What causes positive and negative charge? Ans: Deficiency/removing of electrons in an atom cause positive charge on that atom and excess of electrons in any atom cause –ve charge. III. When two uncharged particles come close to each other, will they repel or attract each other? Ans: When two uncharged particles come close to each other, they do not repel or attract each other. Attraction or repletion occurs in charged particles.