Question

Two identical R = 282ohm light bulbs are connectedin parallel to a 112.6volt battery. How much power(Watt) is supplied by thebattery?

Answer 1

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

`\begin{gathered} R_t=(R+R)/(R.R) \\ =(2R)/(R^2) \\ =(2)/(R) \end{gathered}`

So, substituting the value of R in the above equation we get

`\begin{gathered} R_t=(2)/(282) \\ =7.09\text{ }*10^(-3)\text{ }\Omega \end{gathered}`

Thus the power will be

`\begin{gathered} P=(V^2)/(R_t) \\ =((112.6)^2)/(7.09*10^(-3)) \\ =1.7*10^6\text{ W} \end{gathered}`

Thus the power is 1.7 x10^6 W