Asked by tangocharliee | 10th Nov, 2009, 02:05: PM
There is only one junction which is common to FOUR resistors.
Since the the junction point is a point of symmetry for the circuit i.e. to the left and right of that junction point the resistors and their connection are perfectly identical.
So the current entering the junction is equal to the current leaving. That is the current entering from horizontal resistance from A will leave along the horizontal resistance to B and this will be true for the other two resistances at the junction.
So can disconnect the the resistors at the this point, since the currents flowing remain unchanged.
Then we can see that the resulting circuit has upper half with three resistance in series,
First is 1Ω, second is the equivalent of parallel combination of 1Ω and 1+1 = 2Ω, and the third is again 1Ω.
The equivalent of upper half = 1 + 2/3 + 1 = 8/3 Ω
The lower half is the direct resistances between A and B i.e. 1Ω and 1Ω.
Hence the equivalent resistance = parallel combination of upper half and lower half =
(8/3)(2)/(8/3 + 2) = 8/7 Ω
Answered by | 10th Nov, 2009, 09:11: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number