In the circuit as shown, it is given that `R_1//R_2 = 1//2` and potential at O is V.

Options

A. `150 Omega`
B. `300 Omega`
C. `450 Omega`
D. `600 Omega`

Answer

D

Explanation

d. Given, `R_2 = 2R_1`
For `AO , 100 -V = i_1R_1`
For `BO, 100- V = i_2(2R_2)`
For `OD, V-0 = i_3 (20)`
For `OC , V - 50 = (i_1 + i_2/2 - i_3)(50)`
`VR_1 = 50R_1 = ((3i_1)/2 xx 50 -i_3 xx 50) R_1`
From Eqs. (i) and (iii),
`VR_(1) = 50R_(1) = 75(100-V) -2.5 VR_(1)`
`3.5 VR_1 - 50R_1 = 7500 - 75V` ,brgt If `Vlt75V,` then `RHSgt0`.
So, `LHS gt0`
`3.5 V gt 50 ("since" R_1!=0) `
`Vgt50/3.5`
From Eq. (v), `V=20 V`.
`70 R_1 - 50R_1 = 7500 - 1500 = 6000`
`R_1 = 300 Omega`
`R_2 = 2R_1 = 600Omega`
From Eq. (v),
`3.5 VR_1 + 75V = 7500 + 50 R_1`
For any value of `R_1 Vgt0`.