Question

A wire of `15 Omega` resistance is gradually stretched to double in original length. it is then cut into two equal parts .These parts are then connected in parallel across a `3.0` volt battery. Find the current draw from the battery .

Answer 1

Correct Answer - `0.2 A`
When a resisteor is stretched to double in original length the new resistance becomes four times its original resistance becomes `R prop l^(2)`
`:.` New resistance `= 4 R = 4 xx 15 = 60 Omega`
Resistance of each part `= 30 Omega`
i.e. `R_(1) = 30 Omega and R_(2) = 30 Omega`
When they are conpected in parallel the effective
resistance, `R_(P) = (R_(1)R_(2))/(R_(1) + R_(2)) = (30 xx 30)/( 30 + 30) = 15 Omega`
Current, `I = (V)/(R_(P)) = (3)/(15) = 0.2 A`