Concept:The instantaneous power dissipated in a purely resistive AC circuit is calculated using the value of the instantaneous voltage at that specific moment in time.
Formula:$$V(t) = V_0 \sin(2\pi f t)$$
$$P_{\text{inst}} = \frac{[V(t)]^2}{R}$$
Solution:Given values:
- Peak voltage \( V_0 = 20\text{ V} \)
- Frequency \( f = 15\text{ Hz} \)
- Time \( t = \frac{1}{180}\text{ s} \)
- Resistance \( R = 10\ \Omega \)
1. Calculate the instantaneous voltage:
$$V(t) = 20 \sin\left(2\pi \times 15 \times \frac{1}{180}\right) = 20 \sin\left(\frac{30\pi}{180}\right)$$
$$V(t) = 20 \sin\left(\frac{\pi}{6}\right) = 20 \times 0.5 = 10\text{ V}$$
2. Calculate the instantaneous power:
$$P_{\text{inst}} = \frac{(10\text{ V})^2}{10\ \Omega} = \frac{100}{10} = 10\text{ W}$$
Why other options are incorrect:- 100 W would be the power if we incorrectly used the peak voltage (\( V_0 = 20\text{ V} \)) to calculate maximum power (\( P_{\text{max}} = \frac{20^2}{10} = 40\text{ W} \)) or made a math error.
- 40 W is the absolute maximum power of the AC cycle, not the power at the given instant.
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