The equivalent resistance between the terminals A and B in the network shown in the figure is

Options

A. `7Omega`
B. `14Omega`
C. `15Omega`
D. `21Omega`

Answer

D

Explanation

Apply KVL for the loop CKHDC we have,
`4I_(1)-I_(2)=2I` ...(i)
Apply KVL for the loop DHGFD we have,
`-I_(1)+4I_(2)=I` ... (ii)
From equation (i) and (ii), we have,
`I_(1)=(3)/(5) I ` and `I_(2)=(2)/(3)I`
Apply KVL for the loop ACKHGBEA
we have,
`-30 (I-I_(1))-15(I-I_(2))+E=0`
(After solving this equation and by putting values of `I_(1)` and `I_(2)` in this equations)
`E=(7)/(5)IR` .... (iii)
Let R' be the resistance between A and B
`:.E=IR'` ...(iv)
From equation (iii) and (iv), we have,
`IR' = 7/5 IR`
`:.R' = (7)/(5)R=(7)/(5)xx15=21Omega`.