Question 374: AC fundamentals

Question

The a.c. source is connected to series LCR circuit. If voltage across $$R$$ 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

Accepted Answer

The emf across the $$\mathrm{AC}$$ source is given as:
$$\begin{aligned}
\mathrm{e} & =\sqrt{\left(\mathrm{V}_{\mathrm{R}}ight)^2+\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{L}}ight)^2}=\sqrt{(40)^2+(80-40)^2} \
\mathrm{e} & =\sqrt{3200} \
	herefore \quad \mathrm{e} & =40 \sqrt{2} \mathrm{~V}
\end{aligned}$$

Answer

40 V

Answer

40$$\sqrt2$$ V

Answer

80 V

Answer

160 V

AC fundamentals

The a.c. source is connected to series LCR circuit. If voltage across $R is 40 \mathrm{~V}, that across \mathrm{L} is 80 \mathrm{~V} and that across \mathrm{C} is 40 \mathrm{~V}, then the e.m.f. 'e$' of a.c. source is

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