Gujarat BoardEnglish MediumSTD 12 SciencePhysicsJUNE 20244 Marks
Question
Explain AC circuit with only an inductor.
✓
Answer
→ As shown in fig., an AC source is connected to pure inductor. (A pure inductor means an inductor having negligibly small resistance.) → Let the voltage across the source be $ v=v_m \sin \omega t ...(1)$ → Using the kirchhoff's loop rule for the AC circuit shown in the fig., $ v-L \frac{d i}{d t}=0 ...(2)$ → From equation (2), $ \begin{array}{l} \therefore \quad v=L \frac{d i}{d t} \\ \therefore \quad v_m \sin \omega t=L \frac{d i}{d t} \\ \therefore \quad \frac{d i}{d t}=\frac{v_m}{L} \sin \omega t ...(3)\end{array} $ → To obtain the current, we integrate equation (3) $ \begin{array}{l} \therefore \quad \int \frac{d i}{d t} d t=\int \frac{v_m}{L} \sin (\omega t) d t \\ \therefore \quad i=-\frac{v_m}{\omega L} \cos (\omega t)+\text { constant } \end{array} $ → The integration constant has the dimension of current and is time-independent. Since the source has an emf which oscillates symmetrically about zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. $ \therefore \quad i=-\frac{v_m}{\omega L} \cos \omega t $ $ \therefore \quad i=\frac{v_m}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right) $ $\therefore \quad i=i_m \sin \left(\omega t-\frac{\pi}{2}\right)...(4)$ $i_m=\frac{ U _m}{\omega L}$ Amplitude of the current → The quantity $\omega L$ is analogous to the resistance and is called inductive reactance, denoted by $X _{ L }$ : $ \therefore \quad X_L=\omega L $ Unit of $X _{ L }$ is ohm ( $\Omega$ ). → From eq. (1) and (4), it can be said that current lags the voltage by $\frac{\pi}{2} rad$. [or onequarter $\frac{1}{4}$ cycle].
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