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

Note by Aßhĩмanyu Singh
6 years, 4 months ago

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

(11 Ω+11 Ω+11 Ω)1+(13 Ω+13 Ω+13 Ω+1R+1R+1R)1+(11 Ω+11 Ω+11 Ω)1=1 Ω(\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,

(13 Ω)+(11 Ω+3R)1+(13 Ω)=1 Ω(\frac{1}{3} \ \Omega) + (\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} + (\frac{1}{3} \ \Omega) = 1 \ \Omega

(11 Ω+3R)1=13 Ω(\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} = \frac{1}{3} \ \Omega

11 Ω+3R=31 Ω\frac{1}{1 \ \Omega}+\frac{3}{R}=\frac{3}{1 \ \Omega}

3R=21 Ω\frac{3}{R}=\frac{2}{1 \ \Omega}

R=32 ΩR=\frac{3}{2} \ \Omega.

The following simulation shows that when a 5 V5 \ \mathrm{V} voltage is applied across the circuit, a 5 A5 \ \mathrm{A} current results, which implies the equivalent resistance of the circuit is 1 Ω1 \ \Omega, verifying that R=32 ΩR=\frac{3}{2} \ \Omega is correct: https://imgur.com/WoS0Yn7

Ricky Escobar - 6 years, 4 months ago

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hey ricky is there any other method ????

Ritvik Choudhary - 6 years, 4 months ago

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abe mujhe to ye method jyaada easy lag rha hai

AßHĨМANYU Singh - 6 years, 4 months ago

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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 - 6 years, 4 months ago

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What will be the current in 3 ohm resistance?

Shubham Srivastava - 6 years, 4 months ago

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19 A0.1111 A\frac{1}{9} \ \mathrm{A} \approx 0.1111 \ \mathrm{A}

Ricky Escobar - 6 years, 4 months ago

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plzzz anyone

AßHĨМANYU Singh - 6 years, 4 months ago

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sory where do you get that question

Gonna Sing - 6 years, 4 months ago

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u can use star & delta techniques that will simplify it

Amgad Ahmed - 6 years, 4 months ago

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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 - 6 years, 4 months ago

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hey, actually i am in class 10th and we have not covered this topic yet.

AßHĨМANYU Singh - 6 years, 4 months ago

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although i know abot v , i and parallel and series circuits

AßHĨМANYU Singh - 6 years, 4 months ago

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Try applying Kirchoff's Law. By Conservation of Charge, the current entering at A should leave at B, right? So start there.

Aditya Karekatte - 6 years, 4 months ago

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plzzz can u give me step by step solution to this

AßHĨМANYU Singh - 6 years, 4 months ago

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FIrst Reply.. u are in which standard and do you know Kirchoff Law

A Former Brilliant Member - 6 years, 4 months ago

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@A Former Brilliant Member i m in 10th standard, and according to cbse 10 physics syllabi, there is nothing as kirchoff law

AßHĨМANYU Singh - 6 years, 4 months ago

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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 - 6 years, 4 months ago

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symertrycity will help u calculate the some point as same if distance from source and sink are same it will make easier

Dhurjati Das - 6 years, 4 months ago

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1ohm

Ashutosh Giri - 5 years, 7 months ago

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From SOmewhere i got the hint that it's ans is 4/3

AßHĨМANYU Singh - 6 years, 4 months ago

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is it 3/8?

Mahathir Ahmad - 6 years, 4 months ago

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mantap, dapet soal darimananih?

Putu Andika - 6 years, 4 months ago

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English, please?

Aditya Karekatte - 6 years, 4 months ago

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okay, i'll

Putu Andika - 6 years, 4 months ago

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that is indonesian language. iya mantap nih

Ridhan Fadhilah - 6 years, 4 months ago

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R=3 DUE TO SYMMETRY OF THE CIRCUIT

Satyam Suman - 6 years, 4 months ago

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what do you mean, can't it be solved with kirchoff's law. well i dont know about it

AßHĨМANYU Singh - 6 years, 4 months ago

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