Waste less time on Facebook — follow Brilliant.
×

R Woessss

Can plzzz anyone give me step by step solution to this...

Note by Aßhĩмanyu Singh
3 years, 10 months ago

No vote yet
16 votes

Comments

Sort by:

Top Newest

Using the methods outlined on this page: http://www.rfcafe.com/miscellany/factoids/kirts-cogitations-256.htm, you can set up the following equation:

\((\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega})^{-1} + (\frac{1}{3 \ \Omega}+\frac{1}{3 \ \Omega}+\frac{1}{3 \ \Omega}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R})^{-1} +(\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega})^{-1} = 1 \ \Omega\).

Simplifying,

\((\frac{1}{3} \ \Omega) + (\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} + (\frac{1}{3} \ \Omega) = 1 \ \Omega\)

\((\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} = \frac{1}{3} \ \Omega\)

\(\frac{1}{1 \ \Omega}+\frac{3}{R}=\frac{3}{1 \ \Omega}\)

\(\frac{3}{R}=\frac{2}{1 \ \Omega}\)

\(R=\frac{3}{2} \ \Omega\).

The following simulation shows that when a \(5 \ \mathrm{V}\) voltage is applied across the circuit, a \(5 \ \mathrm{A}\) current results, which implies the equivalent resistance of the circuit is \(1 \ \Omega\), verifying that \(R=\frac{3}{2} \ \Omega\) is correct: http://imgur.com/WoS0Yn7 Ricky Escobar · 3 years, 10 months ago

Log in to reply

@Ricky Escobar hey ricky is there any other method ???? Ritvik Choudhary · 3 years, 9 months ago

Log in to reply

@Ritvik Choudhary abe mujhe to ye method jyaada easy lag rha hai Aßhĩмanyu Singh · 3 years, 9 months ago

Log in to reply

Look at the circuit, It is symmetric for any value of R. We can proceed to result in many ways including Kirchoff's law , One of the easiest method is given here . It can be only applied for a symmetric circuit(caution). Consider a 1 A current is flowing through it , as the circuit is symmetric 1/3 A will flow through each 1 ohm .Also 1/3 A will spit to 3 ohm and R, then finally 1/3 A will flow through last 1 ohm (1/3 A through each 1 ohm). Thus we get over all resistance as 1/3+(3 parallel to R)/3+1/3 =1 ,(3 parallel to R)/3=1/3 ,Thus R=3/2 . If any explanation needed, then please specify Muhammed Shemeer K · 3 years, 10 months ago

Log in to reply

@Muhammed Shemeer K What will be the current in 3 ohm resistance? Shubham Srivastava · 3 years, 10 months ago

Log in to reply

@Shubham Srivastava \(\frac{1}{9} \ \mathrm{A} \approx 0.1111 \ \mathrm{A}\) Ricky Escobar · 3 years, 10 months ago

Log in to reply

1ohm Ashutosh Giri · 3 years ago

Log in to reply

symertrycity will help u calculate the some point as same if distance from source and sink are same it will make easier Dhurjati Das · 3 years, 10 months ago

Log in to reply

hey abhimanyu u can try to unferstand this. as points A and B are given name other 6 points as C,D,....H. Then try to figure out if a current say I starts from point A u can see the just next 3 corners are all connected by a 1 ohm resistor. so the current will be divided equally. then try to figure out those corners and the wires connecting then are they equal? try to find out the symmetry. U also can find from back ward also. after starting from A u try to picturise if the current I enters b and the next three corners of B are all equal resistance wire connected so the current would have come all equally from all those 3 corners try to ratioanalise ur thinking this way. i hope u will solve the remaining part urself. Soham Mullick · 3 years, 10 months ago

Log in to reply

Try applying Kirchoff's Law. By Conservation of Charge, the current entering at A should leave at B, right? So start there. Aditya Karekatte · 3 years, 10 months ago

Log in to reply

@Aditya Karekatte plzzz can u give me step by step solution to this Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

@Aßhĩмanyu Singh FIrst Reply.. u are in which standard and do you know Kirchoff Law Arushit Mudgal · 3 years, 10 months ago

Log in to reply

@Arushit Mudgal i m in 10th standard, and according to cbse 10 physics syllabi, there is nothing as kirchoff law Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

There is a similar question in haliday,resnick and krane volume-2 physics book.Basically,here,you have to notice the symmetry and the points with the same potential.Through it,you can rearrange this circuit and solve it.Hopefully if i have not done any mistake in calculation the answer is 3/8. Mahathir Ahmad · 3 years, 10 months ago

Log in to reply

@Mahathir Ahmad although i know abot v , i and parallel and series circuits Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

@Mahathir Ahmad hey, actually i am in class 10th and we have not covered this topic yet. Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

u can use star & delta techniques that will simplify it Amgad Ahmed · 3 years, 10 months ago

Log in to reply

sory where do you get that question Gonna Sing · 3 years, 10 months ago

Log in to reply

plzzz anyone Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

mantap, dapet soal darimananih? Putu Andika · 3 years, 10 months ago

Log in to reply

@Putu Andika that is indonesian language. iya mantap nih Ridhan Fadhilah · 3 years, 10 months ago

Log in to reply

@Putu Andika English, please? Aditya Karekatte · 3 years, 10 months ago

Log in to reply

@Aditya Karekatte okay, i'll Putu Andika · 3 years, 10 months ago

Log in to reply

is it 3/8? Mahathir Ahmad · 3 years, 10 months ago

Log in to reply

From SOmewhere i got the hint that it's ans is 4/3 Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

R=3 DUE TO SYMMETRY OF THE CIRCUIT Satyam Suman · 3 years, 10 months ago

Log in to reply

@Satyam Suman what do you mean, can't it be solved with kirchoff's law. well i dont know about it Aßhĩмanyu Singh · 3 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...