When three resistors are connected in series then total equivalent of all three is given by
![R_(eq)= R_1 + R_2 + R_3](https://img.qammunity.org/2019/formulas/physics/middle-school/j0zljjqycmsaxph0rtz2jpok8tg1kfe0ly.png)
Also in this case current is given by
![i = \frac{V}{R_(eq)]()
Now if one of the resistance is removed
Then
![R_(eq) = R_1 + R_2](https://img.qammunity.org/2019/formulas/physics/middle-school/c9pxthe67w2fpo7o3vkbjyacfqakvyf4gc.png)
let one of the resistance is removed
while current flowing in the open circuit will be given by
![i = (V)/(R_(eq))](https://img.qammunity.org/2019/formulas/physics/middle-school/36ajg2mi5isj55edh1zk7hg13h4vo8u4xu.png)
So here since resistance is decreased then current must be increased in the circuit.