Question

the following two formulas are used in electricity.

\begin{array}{l} p = i ^ { 2 } r \\ e ^ { 2 } = p r end{array} where ( left{ \begin{array} { l } { p = \text { power (in watts) } } \\ { i = \text { current (in amperes) } } \\ { e = \text { voltage (in volts) } } \\ { r = \text { resistance (in ohms, } omega ) } end{array}
ight. )

a ( 13.000 omega ) resistor is rated at 3.5 watts. what is the maximum voltage of electricity that should be connected across it? round your answer to two decimal places, if necessary.

answer

volts

Explanation:

Step1: Substitute the given values into the formula \(E^{2}=PR\)

Given \(P = 3.5\) watts and \(R=13000\space\Omega\). Then \(E^{2}=3.5\times13000\).

Step2: Calculate the value of \(E^{2}\)

\(E^{2}=3.5\times13000 = 45500\).

Step3: Solve for \(E\)

Take the square - root of both sides: \(E=\sqrt{45500}\). Using a calculator, \(E\approx213.31\) volts.

Answer:

\(213.31\) volts