1. **Stating the problem:** Given two time-dependent functions representing voltage and current in an AC circuit: $$v(t) = 20\cos(4t + 30^\circ)\text{ V}$$ $$i(t) = 70\cos(5t + 60^\circ)\text{ A}$$ We want to analyze these functions, understand their properties, and possibly find related quantities such as instantaneous power. 2. **Formula and important rules:** - The voltage and current are given as cosine functions with angular frequencies and phase shifts. - Angular frequency $\omega$ is the coefficient of $t$ inside the cosine. - Phase angles are given in degrees and should be converted to radians if needed for calculations. - Instantaneous power $p(t)$ in an AC circuit is given by: $$p(t) = v(t) \times i(t)$$ 3. **Intermediate work:** - Identify angular frequencies: - Voltage frequency: $\omega_v = 4$ rad/s - Current frequency: $\omega_i = 5$ rad/s - Phase angles: - Voltage phase: $30^\circ = \frac{\pi}{6}$ radians - Current phase: $60^\circ = \frac{\pi}{3}$ radians 4. **Calculating instantaneous power:** $$p(t) = 20\cos(4t + 30^\circ) \times 70\cos(5t + 60^\circ) = 1400 \cos(4t + 30^\circ) \cos(5t + 60^\circ)$$ Using the product-to-sum identity: $$\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$$ So, $$p(t) = 1400 \times \frac{1}{2} [\cos((4t + 30^\circ) - (5t + 60^\circ)) + \cos((4t + 30^\circ) + (5t + 60^\circ))]$$ $$= 700 [\cos(-t - 30^\circ) + \cos(9t + 90^\circ)]$$ Since $\cos(-x) = \cos x$ and $\cos(\theta + 90^\circ) = -\sin \theta$, we get: $$p(t) = 700 [\cos(t + 30^\circ) - \sin(9t)]$$ 5. **Explanation:** - The voltage and current have different frequencies, so the instantaneous power is a combination of two oscillations: one at frequency 1 rad/s and another at 9 rad/s. - The power oscillates and does not have a constant average value due to the frequency difference. **Final answer:** The instantaneous power is: $$p(t) = 700 \cos(t + 30^\circ) - 700 \sin(9t)$$ This expression shows how power varies over time in this AC circuit with different voltage and current frequencies.