Given that there are two bulbs connected in parallel with resistance, R = 282 ohm each. Also, they are connected to a battery whose voltage, V = 112.6 volt
Here, the formula to find total resistance in parallel is
[tex]\begin{gathered} R_t=\frac{R+R}{R.R} \\ =\frac{2R}{R^2} \\ =\frac{2}{R} \end{gathered}[/tex]So, substituting the value of R in the above equation we get
[tex]\begin{gathered} R_t=\frac{2}{282} \\ =7.09\text{ }\times10^{-3}\text{ }\Omega \end{gathered}[/tex]Thus the power will be
[tex]\begin{gathered} P=\frac{V^2}{R_t} \\ =\frac{(112.6)^2}{7.09\times10^{-3}} \\ =1.7\times10^6\text{ W} \end{gathered}[/tex]Thus the power is 1.7 x10^6 W
