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naz edited Jun 28, 2026, 7:25 PM

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		+	### Условие
		+
		+	$11.4.12.$ [Вставьте условие задачи]
		+
		+	### Решение
		+
		+	### Statement
		+
		+	$11.4.12.$ a. In the vector diagram, the current $I = I_0cosωt$ is defined as the projection onto the $x$-axis of the vector $I_0$, which rotates around the point $O$ with angular velocity $\omega$. How are the voltage drop vectors located on the same diagram when this current flows through the resistance $R$, the inductor $L$ and the capacitance capacitor $C$? What are the amplitudes of the voltage vectors?
		+
		+	b. Using a vector diagram, determine the voltage drop in the center of a series-connected inductor $L$, resistance $R$, and capacitance capacitor $C$ and the phase shift between current and voltage in the circuit if the current in the circuit varies cosinusoidally: $I = I_0coswt$.
		+
		+	![For problem $11.4.12$\|1074x696, 50%](../../img/11.4.12/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-28 115611.png)
		+
		+	### Solution
		+
		+	a) The voltage drop across the resistor is just Ohm's law:
		+
		+	$$V_R = I_0 R cos \omega t$$
		+
		+	There is a charge on the capacitor:
		+
		+	$$q = V_C C= \int_{0}^{t} I(t) dt$$
		+
		+	$$V_C C = \frac{I_0}{\omega} sin(\omega t)$$
		+
		+	$$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$
		+
		+	Finally, we write down the equation for the voltage across the coil:
		+
		+	$$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omega t)$$
		+
		+	$$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omega t + \frac{\pi}{2})$$
		+
		+	The phase shift as seen in the coil $+\frac{\pi}{2}$ depends on the current, in the capacitor $-\frac{\pi}{2}$, the resistance has no shift, the amplitudes are equal:
		+
		+	$$V_R = I_0 R$$
		+	$$V_C = \frac{I_0}{\omega C}$$
		+	$$V_L = LI_0 \omega$$
		+
		+	b)
		+	To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a:
		+
		+	$$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$
		+
		+	Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.:
		+
		+	$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$
		+
		+	#### Answer
		+	a)
		+	$$V_R = I_0 R$$
		+	$$V_C = \frac{I_0}{\omega C}$$
		+	$$V_L = LI_0 \omega$$
		+	![For problem $11.4.12$\|1037x891, 50%](../../img/11.4.12/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-28 115718.png)
		+	b)
		+	$$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$
		+
		+	$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$
		+
		+	#### Ответ
		+
		+	[Вставьте краткий ответ или результат в рамке]

	@@ -0,0 +1,63 @@
			### Условие

			$11.4.12.$ [Вставьте условие задачи]

			### Решение

			### Statement

			$11.4.12.$ a. In the vector diagram, the current $I = I_0cosωt$ is defined as the projection onto the $x$-axis of the vector $I_0$, which rotates around the point $O$ with angular velocity $\omega$. How are the voltage drop vectors located on the same diagram when this current flows through the resistance $R$, the inductor $L$ and the capacitance capacitor $C$? What are the amplitudes of the voltage vectors?

			b. Using a vector diagram, determine the voltage drop in the center of a series-connected inductor $L$, resistance $R$, and capacitance capacitor $C$ and the phase shift between current and voltage in the circuit if the current in the circuit varies cosinusoidally: $I = I_0coswt$.

			![For problem $11.4.12$\|1074x696, 50%](../../img/11.4.12/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-28 115611.png)

			### Solution

			a) The voltage drop across the resistor is just Ohm's law:

			$$V_R = I_0 R cos \omega t$$

			There is a charge on the capacitor:

			$$q = V_C C= \int_{0}^{t} I(t) dt$$

			$$V_C C = \frac{I_0}{\omega} sin(\omega t)$$

			$$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$

			Finally, we write down the equation for the voltage across the coil:

			$$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omega t)$$

			$$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omega t + \frac{\pi}{2})$$

			The phase shift as seen in the coil $+\frac{\pi}{2}$ depends on the current, in the capacitor $-\frac{\pi}{2}$, the resistance has no shift, the amplitudes are equal:

			$$V_R = I_0 R$$
			$$V_C = \frac{I_0}{\omega C}$$
			$$V_L = LI_0 \omega$$

			b)
			To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a:

			$$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$

			Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.:

			$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$

			#### Answer
			a)
			$$V_R = I_0 R$$
			$$V_C = \frac{I_0}{\omega C}$$
			$$V_L = LI_0 \omega$$
			![For problem $11.4.12$\|1037x891, 50%](../../img/11.4.12/Ð¡Ð½Ð¸Ð¼Ð¾Ðº ÑÐºÑÐ°Ð½Ð° 2026-06-28 115718.png)
			b)
			$$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$

			$$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$

			#### Ответ

			[Вставьте краткий ответ или результат в рамке]