In a 10 Ohm resistance is connected to an ac supply v = 100 sin(314t + 37°) V, the power dissipated by the resistance is

Understand the Problem

The question is asking for the power dissipated in a resistor when connected to an alternating current (AC) supply. It provides the resistance value, the voltage equation, and multiple-choice answers. The approach to solve this involves calculating the RMS voltage from the AC voltage equation and using the formula P = V^2/R to find the power.

Answer

The power dissipated by the resistance is $500 \text{ W}$.

Steps to Solve

Identify the RMS Voltage The given voltage equation is $v = 100 \sin(314t + 37^\circ)$. The peak voltage ($V_m$) is 100 V. To find the RMS voltage ($V_{rms}$), use the formula: $$ V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{100}{\sqrt{2}} \approx 70.71 \text{ V} $$
Calculate the Power Dissipated Using the formula for power dissipation in a resistor: $$ P = \frac{V_{rms}^2}{R} $$ Substituting the values: $$ P = \frac{(70.71)^2}{10} = \frac{5000.14}{10} \approx 500 \text{ W} $$

The power dissipated by the resistance is approximately 500 W.

More Information

This result is derived from the basic principles of electrical circuits involving AC power. The RMS voltage is used because it represents the equivalent DC voltage that would deliver the same power to the resistor.

Tips

Misunderstanding the difference between peak and RMS voltage.
Using the peak voltage directly in the power formula without converting it to RMS.

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