Question

$i=15\sin \omega t+20\cos \omega t$.
Alternating current $i$ is passed through a resistance of $8\Omega$. Find the heat developed across R in 10 seconds.

• A. $25\times {10}^3$ J
• B. $125\times {10}^2$ J
• C. $625\times {10}^2$ J
• D. $5\times {10}^4$ J

Answer 1

The correct option is A $25\times {10}^3$ J

$i=15\sin \omega t+20\cos \omega t$
$=25\{\frac{15}{25}\sin \omega t+\frac{20}{25}\cos \omega t\}=25\{\sin \omega t\cos {53}^{\circ }+\cos \omega t\sin {53}^{\circ }\}$
$i=25\sin (\omega t+{53}^{\circ })$
$i_{rms}=\frac{25}{\sqrt{2}}Heat\left(Q\right)=Power\times t=i_{rms}^{2}Rt$
$Q$ $=i_{rms}^{2}Rt$
$={\left(\frac{25}{\sqrt{2}}\right)}^{2}8\times 10=\frac{625}{2}\times 8\times 10$
$=2500\times 10=25\times {10}^{3}J$