Answer:
50W bulb
Explanation:
Considering the relationship between power, resistance and voltage, we know that
[tex]R=\frac {v^{2}}{p}[/tex]
Where R is resistance, V is voltage across the circuit and P is power
Considering 100 W power, its resistance will be
[tex]R_{100}=\frac {v^{2}}{100}[/tex]
Similarly, resistance of a 50 W power will be
[tex]R_{50}=\frac {v^{2}}{50}[/tex]
Clearly, the resistance of 50W is greater than resistance of 100W
For series connection, power will be given by
P=I²R
For 100W bulb, power will be
[tex]P_{100}=I^{2}\frac {v^{2}}{100}[/tex]
For 50W bulb, power will be
[tex]P_{50}=I^{2}\frac {v^{2}}{50}[/tex]
The current will be the same hence it is clear that power of 50W bulb will be more than power of 100W bulb. Therefore, the 50W bulb has more brightness.
