An ac current is represented as

$i=5 \sqrt{2}+10 \cos \left(650 \pi t+\frac{\pi}{6}\right) A m p$

The r.m.s value of the current is

<p>To find the root mean square (RMS) value of the given alternating current, follow these steps:</p> <p>The current is represented as:</p> <p>$ i = 5\sqrt{2} + 10 \cos \left(650\pi t + \frac{\pi}{6}\right) \, \text{Amp} $</p> <p>Here, the time-independent DC component is $5\sqrt{2}$ and the AC component is $10 \cos \left(650\pi t + \frac{\pi}{6}\right)$.</p> <p><p><strong>Calculate the square of the current, $i^2$:</strong></p> <p>$ i^2 = \left(5\sqrt{2}\right)^2 + \left(10 \cos \left(650\pi t + \frac{\pi}{6}\right)\right)^2 + 2 \times 5\sqrt{2} \times 10 \cos \left(650\pi t + \frac{\pi}{6}\right) $</p> <p>Simplifying, we have:</p> <p>$ i^2 = 50 + 100 \cos^2 \left(650\pi t + \frac{\pi}{6}\right) + 100\sqrt{2} \cos \left(650\pi t + \frac{\pi}{6}\right) $</p></p> <p><p><strong>Find the average value $\langle i^2 \rangle$:</strong></p> <p>The average value of $\cos$ terms over a period is zero, simplifying our equation to:</p> <p>$ \langle i^2 \rangle = 50 + \frac{100}{2} + 0 $</p> <p>This simplifies to:</p> <p>$ \langle i^2 \rangle = 50 + 50 = 100 $</p></p> <p><p><strong>Calculate the RMS current:</strong></p> <p>The RMS value is the square root of the mean of the squares of the current:</p> <p>$ \langle i \rangle = \sqrt{100} = 10 \, \text{Amp} $</p></p> <p>Thus, the RMS value of the current is 10 Amps.</p>