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For best illumination of the light bulb L_1, potential across it should be 6V.
The battery in the circuit produces 9V.
So (9-6) = 3V potential has to be dropped by use of resistor `R_1` , which is placed in series with the light bulb.
The voltage at the left end of `R_1` will be 9V and volatge in its right end will be 6V. In consequence there will be a 3V drop in potential across `R_1` , and the light bulb will face the desired voltage of 6V across its two ends.
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