A cell sends a current through a resistance `R_(1)` for time `t`, next the same cell sends current through another resistance `R_(2)` for the time `t` If the same amount of heat is developed in both the resistance then find the internal resistance of the cell

Options

A. `(R_1 R_2)/(R_1 + R_2)`
B. `R_1 + R_2`
C. zero
D. `sqrt(R_1 R_2)`

Answer

D

Explanation

Let internal resistance of source = R
Current in coil of resistance `R_1= I_(1) = (V)/( R+ R_(1))`
Current in coil of resistance `R_(2) = I_(2) = (V)/( R+ R_(2))`
Further, as heat generated is same, so
`I_(1)^(2) R_(1) t= I_(2)^(2) R_(2) t`
or `((V)/(R+ R_1)) R_1 = ((V)/( R+ R_(2))^(2))^(2) R_(2)`
`rArr R_(1) ( R+ R_(2) )^(2) = R_(2) ( R+ R_(1) )^(2)`
`rArr R^(2) R_(1) + R_(1) R_(2)^(2) + 2 "RR"_(1) R_(2)`
`= R^(2) R_(2)+ R_(1)^(2) R_(2) + 2 RR_(1) R_(2)`
`rArr R^(2) (R_(1) - R_(2) )= R_(1) R_(2) (R_(1) - R_(2) ) rArr R= sqrt(R_1 R_2)`