Two batteries `V_(1)` and `V_(2)` are connected to three resistors as shown below. If `V_(1) = 2V `and `V_(2)` = 0 V, the current I = 3 mA. If `V_(1)` = 0 V and `V_(2)` = 4V, the current I = 4mA. Now, if `V_(1)` = 10 V and `V_(2)` = 10 V, the current I will be-

Options

A. 7 mA
B. 15 mA
C. 20 mA
D. 25 mA

Answer

D

Explanation

`v_(eq) = (V_(1)/R_(1) + V_(1)/R_(1))/(1/R_(1) + 1/R_(2))`
`R_(eq) =(R_(1)R_(2))/(R_(1) + R_(2))`
`I= V_(eq)/(R + R_(eq))`

Case (i) `V_(1) = 2, V_(2) =0`
`V_(eq) = (2R_(2))/(R_(1) + R_(2)), I_(1) =(2R_(2))/(R_(1) + R_(2)) 1/(R +(R_(1)R_(2))/(R_(1) +R_(2)))`
Case (ii) `V_(1) = 0,V_(2) = 4`
`V_(eq) = (4R_(1))/(R_(1) + R_(2)), I_(2) = (4R_(1))/(R_(1) + R_(2)) 1/(R + (R_(1)R_(2))/(R_(1) + R_(2)))`
Now, `I_(1)/I_(2) = 3/4 rArr 3/4 = (2R_(2))/(4R_(1)) rArr R_(2)/R_(1) = 3/2`
Case (iii) `V_(1) = 10 V` and `V_(2) = 10 V`
`V_(eq) = (10 R_(1) + 10R_(2))/(R_(1) + R_(2))`
`I. = (10(R_(1) + R_(2)))/(R_(1) + R_(2)) .1/(R + (R_(1)R_(2))/(R_(1)+R_(2)))`
`I_(1)/I = 3/I^(.) = (2R_(2))/(10 (R_(1) + R_(2))) rArr 3/I^(.) = (2 xx 1.5 R_(1))/(10(2.5 R_(1))`
`rArr I^(.) =25 mA`