Question

A light bulb can’t exceed 0.02A, else it will burn out. If it's connected in series to these resistors (100Ω, 15Ω, 30Ω, 310Ω, xΩ) on a 9V battery, what is the approximate value of the 'x' resistor in the circuit that will produce the maximum current through the bulb ?

Answer 1

By Ohm's law, the current is

$$I=\frac VR=\frac V{R_1+R_2+R_3+R_4+R_x},\\\space\\ R_x=\frac VI-(R_1+R_2+R_3+R_4)=0\Omega$$

because the equivalent resistance of the 4 resistors already gives a current of 0.0198 A.