Answer:False
Explanation:
Given
Power of first lamp [tex]P=50\ W[/tex] at [tex]V=120\ V[/tex]
Power of second lamp [tex]P=75\ W[/tex] at [tex]V=120\ V[/tex]
Resistance of two lamps is given by
[tex]R=\frac{V^2}{P}[/tex]
[tex]R_1=\frac{120^2}{50}[/tex]
[tex]R_1=288\ \Omega[/tex]
For [tex]R_2=\frac{120^2}{75}[/tex]
[tex]R_2=192\ \Omega[/tex]
If both the lamps are connected in series with 240 V source then the current in the circuit is
[tex]i=\frac{V}{R}[/tex]
[tex]i=\frac{240}{192+288}[/tex]
[tex]i=0.5\ A[/tex]
Power of first lamp [tex]P_1=i^2R_1[/tex]
[tex]P_1=(0.5)^2\times 288[/tex]
[tex]P_1=72\ W[/tex]
[tex]P_2=(0.5)^2\times 192[/tex]
[tex]P_2=48\ W[/tex]
Thus both operate at lower range of rated power
