Question 3: AC fundamentals

Question

In a $$L C R$$ series circuit source voltage is $$120 \mathrm{~V}$$ and voltage in inductor $$50 \mathrm{~V}$$ and resistance is $$40 \mathrm{~V}$$, then determine voltage in the capacitor

Accepted Answer

Given, in $$L C R$$ series circuit,
Source voltage (resultant voltage) $$V=120 \mathrm{~V}$$
Voltage across inductor, $$V_L=50 \mathrm{~V}$$
Voltage across resistor, $$V_R=40 \mathrm{~V}$$
In $$L$$-$$C$$-$$R$$ series circuit,
$$\begin{aligned}
V & =\sqrt{V_R^2+\left(V_L-V_Cight)^2} \
	ext{or} \quad 120 & =\sqrt{40^2+\left(50-V_Cight)^2} \
120^2 & =40^2+\left(50-V_Cight)^2
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow \quad\left(50-V_Cight)^2=12800 \
& \Rightarrow \quad 50-V_C=80 \sqrt{2} \Rightarrow V_C=10(5-8 \sqrt{2})
\end{aligned}$$

Answer

$$V_C=10(5-8 \sqrt{2})$$

Answer

$$V_C=10(5+8 \sqrt{2})$$

Answer

$$V_C=20(5+8 \sqrt{2})$$

Answer

$$V_C=10(5+7 \sqrt{2})$$

AC fundamentals

In a $L C R series circuit source voltage is 120 \mathrm{~V} and voltage in inductor 50 \mathrm{~V} and resistance is 40 \mathrm{~V}$, then determine voltage in the capacitor

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