Consider the circuit diagram as shown in the figure. The source has a voltage $V=V_0 \sin \omega t$....

The term $\frac{j}{\omega C}$ represents the impedance of the capacitor in the circuit. In AC circuits, capacitive impedance is given by $Z_C=\frac{j}{\omega C}$ where $j=\sqrt{-1}$ and $\omega$ is the angular frequency of the source.Similarly,$$ Z_L=j \omega L $$We'll apply Kirchhoff's Current Law (KCL) to both loops of the circuit. By analyzing the phasor (complex) impedances and equating real and imaginary parts, we determine the condition for in-phase current.KCL in loop 1:$$ \begin{gathered} \left.i_2 R-\frac{i_2 j}{\omega C}-i_1 R=0 \quad \text { (equation } 1\right) \\ i_1 R=i_2\left(R-\frac{j}{\omega C}\right) \\ i_1=i_2\left(1-\frac{j}{\omega R C}\right) \end{gathered} $$KCL in loop 2:$$ \begin{gathered} i_2 R+i_2\left(\frac{-j}{\omega C}\right)+\left(i_1+i_2\right) j \omega L=V_{\text {avg }}=0 \\ i_2\left(R-\frac{j}{\omega C}\right)=\left(i_1+i_2\right) j \omega L \end{gathered} $$On substituting,$$ i_2\left(R-\frac{j}{\omega C}\right)=\left(i_2\left(1-\frac{j}{\omega R C}\right)+i_2\right) j \omega L $$Cancelling out $i_2$,$$ R-\frac{j}{\omega C}=\left(2-\frac{j}{\omega R C}\right) j \omega L $$Since $j=\sqrt{-1}, j^2=-1$$$ R-\frac{j}{\omega C}=2 j \omega L+\frac{L}{R C} $$In this context, 'in phase' means that the voltage and current waveforms align perfectly over time, without any phase shift between them. For this to happen in an AC circuit, the imaginary (reactive) components of the circuit's impedance must cancel each other out. When this cancellation occurs, the circuit behaves as if it only has resistive components, leading to the current being in sync with the voltage$$ \begin{gathered} \Rightarrow \frac{j}{\omega C}=2 j \omega L \\ \omega^2=\frac{1}{2 L C} \\ \omega=\frac{1}{\sqrt{2 L C}} \end{gathered} $$

Alternating Current - IAT-IISER Physics PYQs

The term

%5Cfrac%7Bj%7D%7B%5Comega%20C%7D

represents the impedance of the capacitor in the circuit. In AC circuits, capacitive impedance is given by

Z_C%3D%5Cfrac%7Bj%7D%7B%5Comega%20C%7D

where

j%3D%5Csqrt%7B-1%7D

and

%5Comega

is the angular frequency of the source.Similarly,$

Z_L%3Dj%20%5Comega%20L

We'll%20apply%20Kirchhoff's%20Current%20Law%20(KCL)%20to%20both%20loops%20of%20the%20circuit.%20By%20analyzing%20the%20phasor%20(complex)%20impedances%20and%20equating%20real%20and%20imaginary%20parts%2C%20we%20determine%20the%20condition%20for%20in-phase%20current.KCL%20in%20loop%201%3A

%5Cbegin%7Bgathered%7D%20%5Cleft.i_2%20R-%5Cfrac%7Bi_2%20j%7D%7B%5Comega%20C%7D-i_1%20R%3D0%20%5Cquad%20%5Ctext%20%7B%20(equation%20%7D%201%5Cright)%20%5C%5C%20i_1%20R%3Di_2%5Cleft(R-%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%5Cright)%20%5C%5C%20i_1%3Di_2%5Cleft(1-%5Cfrac%7Bj%7D%7B%5Comega%20R%20C%7D%5Cright)%20%5Cend%7Bgathered%7D

KCL%20in%20loop%202%3A

%5Cbegin%7Bgathered%7D%20i_2%20R%2Bi_2%5Cleft(%5Cfrac%7B-j%7D%7B%5Comega%20C%7D%5Cright)%2B%5Cleft(i_1%2Bi_2%5Cright)%20j%20%5Comega%20L%3DV_%7B%5Ctext%20%7Bavg%20%7D%7D%3D0%20%5C%5C%20i_2%5Cleft(R-%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%5Cright)%3D%5Cleft(i_1%2Bi_2%5Cright)%20j%20%5Comega%20L%20%5Cend%7Bgathered%7D

On%20substituting%2C

i_2%5Cleft(R-%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%5Cright)%3D%5Cleft(i_2%5Cleft(1-%5Cfrac%7Bj%7D%7B%5Comega%20R%20C%7D%5Cright)%2Bi_2%5Cright)%20j%20%5Comega%20L

Cancelling%20out

i_2

%2C

R-%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%3D%5Cleft(2-%5Cfrac%7Bj%7D%7B%5Comega%20R%20C%7D%5Cright)%20j%20%5Comega%20L

Since

j=\sqrt{-1}, j^2=-1$$

R-%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%3D2%20j%20%5Comega%20L%2B%5Cfrac%7BL%7D%7BR%20C%7D

In%20this%20context%2C%20'in%20phase'%20means%20that%20the%20voltage%20and%20current%20waveforms%20align%20perfectly%20over%20time%2C%20without%20any%20phase%20shift%20between%20them.%20For%20this%20to%20happen%20in%20an%20AC%20circuit%2C%20the%20imaginary%20(reactive)%20components%20of%20the%20circuit's%20impedance%20must%20cancel%20each%20other%20out.%20When%20this%20cancellation%20occurs%2C%20the%20circuit%20behaves%20as%20if%20it%20only%20has%20resistive%20components%2C%20leading%20to%20the%20current%20being%20in%20sync%20with%20the%20voltage

%5Cbegin%7Bgathered%7D%20%5CRightarrow%20%5Cfrac%7Bj%7D%7B%5Comega%20C%7D%3D2%20j%20%5Comega%20L%20%5C%5C%20%5Comega%5E2%3D%5Cfrac%7B1%7D%7B2%20L%20C%7D%20%5C%5C%20%5Comega%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%20L%20C%7D%7D%20%5Cend%7Bgathered%7D

Alternating Current

High-Yield Trend

Questions
1 MCQs

Official Solution

Alternating Current

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs