Question

You obtain a 100 W light bulb and a 50 W light bulb. Instead of connecting them in the normal way, you devise a circuit that places them in series across normal household voltage. Which bulb glows more brightly?

Answer 1

50W bulb

Step-by-step explanation:

Considering the relationship between power, resistance and voltage, we know that

`R=\frac {v^(2)}{p}`

Where R is resistance, V is voltage across the circuit and P is power

Considering 100 W power, its resistance will be

`R_(100)=\frac {v^(2)}{100}`

Similarly, resistance of a 50 W power will be

`R_(50)=\frac {v^(2)}{50}`

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

`P_(100)=I^(2)\frac {v^(2)}{100}`

For 50W bulb, power will be

`P_(50)=I^(2)\frac {v^(2)}{50}`

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.