91Ó°ÊÓ

An AC voltage source typically delivers the following amount of power to a solely resistive load:\(\begin{align}{c}{P_{avg}} = {I_{rms}}\Delta {V_{rms}}...(1)\\{P_{avg}} = \frac{1}{2}{I_{\max }}\Delta {V_{\max }}\end{align}\)

where\({I_{\max }}\)is the peak current,\(\Delta {V_{\max }}\)is the peak voltage,\({I_{rms}}\)is the rms current,\(\Delta {V_{rms}}\)is the rms voltage, and so forth.

In an AC circuit where the current fluctuates sinusoidally,\(\begin{align}{c}{I_{rms}} = \frac{{{I_{\max }}}}{{\sqrt 2 }}....(2)\\{I_{rms}} = 0.707{I_{\max }}\end{align}\)represents the rms current.

where\({I_{\max }}\)denotes the circuit's maximum current.

• Power running through the room:
• Voltage of AC power:

The average power consumption of the room heater is found from equation \((1)\):

\({P_{avg}} = {I_{rms}}\Delta {V_{rms}}\)

Solving for\({I_{rms}}\)we get:

\({I_{rms}} = \frac{{{P_{avg}}}}{{\Delta {V_{rms}}}}\)

Entering the values for\({P_{avg}}\)and\(\Delta {V_{rms}}\)we obtain:

\(\begin{align}{}{I_{rms}} &= \frac{{500\;W}}{{120\;V}}\\{I_{rms}} &= 4.17\;A\end{align}\)

The peak current is found by solving Equation\((2)\)for\({I_{\max }}\):

\(\begin{align}{}{I_{\max }} &= \sqrt 2 {I_{rms}}\\{I_{\max }} &= \sqrt 2 (4.17\;A)\\{I_{\max }} &= 5.89\;A\end{align}\)

Therefore, the current value is obtained as \(5.89\;A\).