We have been working very hard since 2009 to facilitate in learning Read More. We can't keep up without your support. Donate.


Assignment 1(Spring 2013)                           

Circuit Theory (Phy301)

Marks: 25

Due Date: April 29, 2013


DON’T MISS THESE Important instructions:


  • To solve this assignment, you should have good command over first 6 lectures.        
  • Upload assignments properly through LMS, (No Assignment will be accepted through email).
  •  Write your ID on the top of your solution file.      
  • All students are directed to use the font and style of text as is used in this document.
  • Don’t use colorful back grounds in your solution files.    
  • Use Math Type or Equation Editor etc for mathematical symbols.   
  • This is not a group assignment, it is an individual assignment so be careful and avoid copying others’ work. If some assignment is found to be copy of some other, both will be awarded zero marks. It also suggests you to keep your assignment safe from others. No excuse will be accepted by anyone if found to be copying or letting others copy.
  • Don’t wait for the last date to submit your assignment.
  • You can draw circuit diagrams in “Paint” or in “Corel Draw”. The simple and easy way is to copy the given image in Paint and do the required changes in it.    


Q 1:   

Find the equivalent resistance RT of given circuit. Write each step of the calculation to get maximum marks. Draw the circuit diagram of each step otherwise you will loose marks.  














Q 2:

Determine the voltage and current across each resistor for the given circuit. Mention the units of calculated value. 







Q 3:

      Answer the following questions.                                                                                                        


  1.               I.      Why a crow sitting on high voltage bare line, does not get electric shock?         


  1.            II.       A 100 Watt light bulb operates on 140V.How much current does it require?   


  1.          III.      Write the names of different instruments/devices used to measure the voltage and current across a circuit element & way know to connect them.         








Views: 2483


Replies to This Discussion

Sorry for very late reply. (Load sheading Zindabaad). 1st, draw equivalent circuit having source voltage in series with 4k and 2k resistance. Then apply voltage division rule to find voltage across both the resistors. Now 4k will remain un-touched, but replace the equivalent 2k resistance with parallel combination of 3k and 6k. Voltage across both of them will be the same as they are in parallel.  Finally apply Ohm's law to find current.

current 4 that three resistance will be same as they said I1 , I1 I1 repeatedly

then ap ny 1.33 and 0.66 kaisy nikala?

diagram b draw kri hy y just formula put krna hy?

logical bta do

current passing through 4k will be 2mA, but when the current reaches the point where there starts the parallel combination of 3k and 6k, current will divide. Isn't it??? We are asked to calculated current across EACH resistance, not total current, and also if we are given current source then same current will be passed through series combination of 4k and 2k. We have already calculated voltage across each, then why can't Ohm's Law is not valid? If you say that same current (2mA) is passing through all of them, then Ohm's Law fails on 3k and 6k resistance. Isn't It??????

plz tell me 0.66 n 1.33 kaisy nikala n kiun?

V = IR, I = v / R, I = 8/4k = 2mA. Just like this

1st question = 9.58 ohms

2nd question: 8 volts across 4k and 4 volts across both 3k and 6k. Currents = 2mA, 1.33mA and 0.66mA respectively.

3rd question = b part: 2.5 amperes.




This website shows the replies in linear order. It's not fair!! I've replied to Maham naeem and my post is shown at the bottom of the page. New posts must be on top of the threads. Anyways, today is grace day, don't worry, In 2nd question, 1st draw equivalent circuit, having voltage source in series with 4k and 2k. Now current passing through 4k is 8 / 4k = 2mA. According to you, same current is passing through 2k. OK fine, 2mA through 4k and 2mA through 2k. BUT when you replace 2k by parallel combination of 3k and 6k, current will divide, isn't it?, so that current through 3k plus current through 6k must equal to 2mA. Now, by Ohm's law, after calculating the currents, 1.33mA plus 0.66mA = 1.99mA, isn't it??? Tell me if I'm wrong.

Ist question = 9.58ohms

3rd question = 0.71 A

for question 1 answer check page number 6 of lect # 3


© 2021   Created by + M.Tariq Malik.   Powered by

Promote Us  |  Report an Issue  |  Privacy Policy  |  Terms of Service