Question

A heater coil connected to a 240-Vᵣₘₛ ac line has a resistance of 32Ω. What are the maximum and minimum values of the instantaneous power?

Accepted Answer

Step 1: Understand the relationship between the root mean square (RMS) voltage and the peak voltage in an AC circuit. The peak voltage $$ V_{	ext{peak}}  is $$related to the RMS voltage $$ V_{	ext{rms}}  by $$the equation $$ V_{	ext{peak}} = \sqrt{2} \cdot V_{	ext{rms}} . $$Use this formula to calculate the peak voltage. Step 2: Recall the formula for instantaneous power in an AC circuit: $$ P(t) = \frac{V^2(t)}{R} $$, where $$ V(t)  is $$the instantaneous voltage and $$ R  is $$the resistance. The instantaneous voltage $$ V(t) $$ varies sinusoidally as $$ V(t) = V_{	ext{peak}} \cdot \sin(\omega t) $$, where $$ \omega  is $$the angular frequency. Step 3: Substitute $$ V(t) = V_{	ext{peak}} \cdot \sin(\omega t) $$ into the power formula $$ P(t) = \frac{V^2(t)}{R} . $$This gives $$ P(t) = \frac{(V_{	ext{peak}} \cdot \sin(\omega t))^2}{R} = \frac{V_{	ext{peak}}^2}{R} \cdot \sin^2(\omega t) . $$Step 4: Determine the maximum and minimum values of $$ P(t) . $$The maximum value of $$ \sin^2(\omega t)  is 1$$, and the minimum value is 0. Therefore, the maximum instantaneous power is $$ P_{	ext{max}} = \frac{V_{	ext{peak}}^2}{R} $$, and the minimum instantaneous power is $$ P_{	ext{min}} = 0 . $$Step 5: Use the value of $$ V_{	ext{peak}} $$ calculated in Step 1 and the given resistance $$ R = 32 \ \Omega  to $$express $$ P_{	ext{max}}  as$$ $$ P_{	ext{max}} = \frac{(\sqrt{2} \cdot V_{	ext{rms}})^2}{R} = \frac{2 \cdot V_{	ext{rms}}^2}{R} . $$Substitute $$ V_{	ext{rms}} = 240 \ 	ext{V}  to $$find the numerical value of $$ P_{	ext{max}} .$$

A heater coil connected to a 240-Vᵣₘₛ ac line has a resistance of 32Ω. What are the maximum and minimum values of the instantaneous power?

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Key Concepts

Ohm's Law

Instantaneous Power

RMS Voltage