Edits to “Statement”, “Solution”, “Answer”
en/11.4.12.md
+18 −7
| @@ -1,12 +1,16 @@ | |||
| ### Statement | |||
| − | $11.4.12.$ [Insert the problem 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: $I = I_0coswt$. | ||
| + | |||
| + |  | ||
| + | |||
| ### Solution | |||
| a) The voltage drop across the resistor is just Ohm's law: | |||
| − | $$V_R = I_0 R cos \omegat$$ | ||
| + | $$V_R = I_0 R cos \omega t$$ | ||
| There is a charge on the capacitor: | |||
| $$q = V_C C= \int_{0}^{t} I(t) dt$$ | |||
| $$V_C C = \frac{I_0}{\omega} sin(\omega t)$$ | |||
| $$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$ | |||
| @@ -18,9 +22,9 @@Solution | |||
| Finally, we write down the equation for the voltage across the coil: | |||
| − | $$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omegat)$$ | ||
| + | $$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omega t)$$ | ||
| − | $$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omegat + \frac{\pi}{2})$$ | ||
| + | $$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omega t + \frac{\pi}{2})$$ | ||
| 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: | |||
| $$V_R = I_0 R$$ | |||
| $$V_C = \frac{I_0}{\omega C}$$ | |||
| $$V_L = LI_0 \omega$$ | |||
| @@ -31,12 +35,19 @@Solution | |||
| b) | |||
| To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a: | |||
| − | $$V = I_0 \sqrt{R^2 + (\omegaL - \frac{1}{\omega C})^2}$$ | ||
| + | $$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$ | ||
| Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.: | |||
| − | $$\varphi = arctg \frac{\omegaL - \frac{1}{\omega C}{R}$$ | ||
| + | $$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$ | ||
| #### Answer | |||
| + | a) | ||
| + | $$V_R = I_0 R$$ | ||
| + | $$V_C = \frac{I_0}{\omega C}$$ | ||
| + | $$V_L = LI_0 \omega$$ | ||
| + |  | ||
| + | b) | ||
| + | $$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$ | ||
| − | [Insert a concise answer or boxed result] | ||
| + | $$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$ | ||
| @@ -1,12 +1,16 @@ | |||
| ### Statement | ### Statement | ||
| $11.4.12.$ [Insert the problem 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: $I = I_0coswt$. | |||
|  | |||
| ### Solution | ### Solution | ||
| a) The voltage drop across the resistor is just Ohm's law: | a) The voltage drop across the resistor is just Ohm's law: | ||
| $$V_R = I_0 R cos \omegat$$ | $$V_R = I_0 R cos \omega t$$ | ||
| There is a charge on the capacitor: | There is a charge on the capacitor: | ||
| $$q = V_C C= \int_{0}^{t} I(t) dt$$ | $$q = V_C C= \int_{0}^{t} I(t) dt$$ | ||
| $$V_C C = \frac{I_0}{\omega} sin(\omega t)$$ | $$V_C C = \frac{I_0}{\omega} sin(\omega t)$$ | ||
| $$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$ | $$V_C = \frac{I_0}{\omega C} cos(\omega t - \frac{\pi}{2})$$ | ||
| @@ -18,9 +22,9 @@Solution | |||
| Finally, we write down the equation for the voltage across the coil: | Finally, we write down the equation for the voltage across the coil: | ||
| $$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omegat)$$ | $$V_L = L \frac{dI}{dt} = -LI_0 \omega sin(\omega t)$$ | ||
| $$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omegat + \frac{\pi}{2})$$ | $$V_L = \frac{dI}{dt} = LI_0 \omega cos(\omega t + \frac{\pi}{2})$$ | ||
| 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: | 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: | ||
| $$V_R = I_0 R$$ | $$V_R = I_0 R$$ | ||
| $$V_C = \frac{I_0}{\omega C}$$ | $$V_C = \frac{I_0}{\omega C}$$ | ||
| $$V_L = LI_0 \omega$$ | $$V_L = LI_0 \omega$$ | ||
| @@ -31,12 +35,19 @@Solution | |||
| b) | b) | ||
| To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a: | To find the amplitude, it is enough to add up our voltages vector based on the diagram obtained in a: | ||
| $$V = I_0 \sqrt{R^2 + (\omegaL - \frac{1}{\omega C})^2}$$ | $$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$ | ||
| Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.: | Since we have a right triangle formed from vectors, the phase shift is easy to express in terms of a tangent.: | ||
| $$\varphi = arctg \frac{\omegaL - \frac{1}{\omega C}{R}$$ | $$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$ | ||
| #### Answer | #### Answer | ||
| a) | |||
| $$V_R = I_0 R$$ | |||
| $$V_C = \frac{I_0}{\omega C}$$ | |||
| $$V_L = LI_0 \omega$$ | |||
|  | |||
| b) | |||
| $$V = I_0 \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$$ | |||
| [Insert a concise answer or boxed result] | $$\varphi = arctg \frac{\omega L - \frac{1}{\omega C}}{R}$$ | ||