The a.c. source is connected to series LCR circuit. If voltage across $\mathrm{R}$ is $40 \mathrm{~V}$, that…
The a.c. source is connected to series LCR circuit. If voltage across $\mathrm{R}$ is $40 \mathrm{~V}$, that across $\mathrm{L}$ is $80 \mathrm{~V}$ and that across $\mathrm{C}$ is $40 \mathrm{~V}$, then the e.m.f. 'e' of a.c. source is
$40 \mathrm{~V}$
$40 \sqrt{2} \mathrm{~V}$
$80 \mathrm{~V}$
$160 \mathrm{~V}$
Solution
The emf across the AC source is given as:
$\begin{aligned}
\mathrm{e} & =\sqrt{\left(\mathrm{V}_{\mathrm{K}}\right)^2+\left(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{L}}\right)^2}=\sqrt{(40)^2+(80-40)^2} \\
\mathrm{e} & =\sqrt{3200} \\
\therefore \quad \mathrm{e} & =40 \sqrt{2} \mathrm{~V}
\end{aligned}$