Answer:
Power will be 166.66 watt when the heater are connected in series
Explanation:
We have given power of the first heater [tex]P_1[/tex] = 490 watt
And power of the second heater [tex]P_2[/tex] = 250 watt
We know that in series connection potential difference will be same
So let the potential difference is v
Power is given by [tex]P=\frac{V^2}{R}[/tex]
So [tex]490=\frac{V^2}{R_1}[/tex]
[tex]R_1=\frac{V^2}{490}=0.0020V^2[/tex]
Similarly [tex]R_2=\frac{V^2}{250}=0.04V^2[/tex]
As resistances are connected in series so equivalent resistance [tex]R=R_1+R_2=0.0020V^2+0.04V^2=0.006V^2[/tex]
So power [tex]P=\frac{V^2}{R}=\frac{V^2}{0.006V^2}=166.66Watt[/tex]
