Parallel Connection of Coupled Coils MCQ Quiz - Objective Question with Answer for Parallel Connection of Coupled Coils - Download Free PDF

The correct answer is 'CADBE'.

 Key Points

• First statement is C because it is introductory statement.
• Second statement is A because it is taking about the virtue.
• Third statement is D because impressed will be followed by me.
• Fourth statement is B because greatly and deeply is used for the virtue of Indian people.
• Fifth statement is E lastly.

Therefore, the correct answer is 'Option 2'.

The correct sequence is - One Indian virtue that has impressed me greatly and deeply is the Indian people's freedom and rancour.

The correct answer is 'CADBE'.

 Key Points

• First statement is C because it is introductory statement.
• Second statement is A because it is taking about the virtue.
• Third statement is D because impressed will be followed by me.
• Fourth statement is B because greatly and deeply is used for the virtue of Indian people.
• Fifth statement is E lastly.

Therefore, the correct answer is 'Option 2'.

The correct sequence is - One Indian virtue that has impressed me greatly and deeply is the Indian people's freedom and rancour.

$\begin{array}{l}\Rightarrow {\mathrm{L}}_{\mathrm{e}\mathrm{q}}={\mathrm{L}}_{\mathrm{X}\mathrm{Y}}=\frac{5}{8}+1+1+\left(-2\times \frac{1}{2}\right)\\ \Rightarrow {\mathrm{L}}_{\mathrm{e}\mathrm{q}}=\frac{13}{8}\mathrm{H}=1.625\mathrm{H}\end{array}$

$V_{in}=V\sin \left(\frac{t}{\sqrt{MC}}\right)$

Let the current in the circuit be i₁

_{$\therefore V_0=M\frac{di}{dt}+i_1\times X_c$}

_{$i_1=\frac{V_{in}}{X_L+X_c}$}

Taking Laplace Transform on both sides:

$V_0\left(s\right)=\left(sM+\frac{1}{sC}\right)i_1\left(s\right)$

$=\left(\frac{s^{2}MC+1}{sC}\right)i_1\left(s\right)$

$i_1\left(s\right)=\frac{V\sqrt{MC}}{\left(s^{2}MC+1\right)\left(sL+\frac{1}{sC}\right)}$

Putting i₁(S) in equation (1)

$V_0\left(s\right)=\frac{V\sqrt{MC}}{\left(s^{2}MC+1\right)\left(s^{2}LC+1\right)}\frac{\left(s^{2}MC+1\right)}{sC}$

$V_0\left(s\right)=\frac{V\sqrt{MC}}{s_{LC}^2+1}=V\sqrt{\frac{M}{L}}\frac{\frac{1}{\sqrt{LC}}}{s^2+\frac{1}{LC}}$

Taking inverse Laplace

$V_0\left(t\right)=V\sqrt{\frac{M}{L}}\sin \left(\frac{t}{\sqrt{LC}}\right)$

$\frac{d{V}_0\left(t\right)}{dt}=\frac{V}{L}\sqrt{\frac{M}{C}}\cos \left(\frac{t}{\sqrt{LC}}\right)$

$\frac{d{V}_0\left(0^{+}\right)}{dt}=\frac{V}{L}\sqrt{\frac{M}{C}}$

The given circuit can be transformed to

$\Rightarrow L_{eq}=3+\left(\frac{-8}{7}\right)=\frac{13}{7}H$

The correct answer is 'CADBE'.

 Key Points

• First statement is C because it is introductory statement.
• Second statement is A because it is taking about the virtue.
• Third statement is D because impressed will be followed by me.
• Fourth statement is B because greatly and deeply is used for the virtue of Indian people.
• Fifth statement is E lastly.

Therefore, the correct answer is 'Option 2'.

The correct sequence is - One Indian virtue that has impressed me greatly and deeply is the Indian people's freedom and rancour.

$V_{AB}=\frac{50d\left(i_1-i_2\right)}{dt}-10\frac{d{i}_3}{dt}-10\frac{d\left(i_2-i_3\right)}{dt}$ ----(1)

$\frac{30d\left(i_2-i_3\right)}{dt}-\frac{50d\left(i_1-i_2\right)}{dt}-\frac{10d\left(i_1-i_2\right)}{dt}+\frac{20d{i}_3}{dt}=0$ ---(2)

$\frac{20d{i}_3}{dt}-\frac{30d\left(i_2-i_3\right)}{dt}+\frac{20d\left(i_2-i_3\right)}{dt}-\frac{10d\left(i_1-i_2\right)}{dt}=0$ ---(3)

Let, $\frac{d{i}_1}{dt}=x;\frac{d{i}_2}{dt}=y;\frac{d{i}_3}{dt}=z$

V_AB = 50 x – 50 y – 10 z – 10 y + 10 z

0 = 30 y – 30 z – 50 x + 50 y – 10 x + 10 y + 20 z

0 = 20 z – 30 y + 30 z + 20 y – 20 z – 10 x + 10 y

V_AB = 50 x – 60 y

0 = -60 x + 90 y – 10 z

Or,

60 x = 90 y – 10 z ----(4)

10 x = 0 y + 30 z ----(5)

Solving (4) and (5) for y

190 x = 270 y

$\begin{array}{l}y=\frac{190}{270}x\\ \therefore V_{AB}=50x-60\left(\frac{190}{270}x\right)\end{array}$  

$\begin{array}{l}=7.78\frac{d{i}_1}{dt}\\ \Rightarrow V_{AB}=L_{eq}\frac{di}{dt}\end{array}$  

$\begin{array}{l}{V}_1=-\frac{j{I}_1}{2C}+j4I_1+j2\sqrt{2}{I}_2\\ 0=\left(4+4j\right)I_2+j2\sqrt{2}{I}_1\\ \Rightarrow I_2=\frac{-j\sqrt{2}{I}_1}{2\left(1+j\right)}\\ \frac{V_1}{I_1}=\frac{-j}{2C}+j4+\frac{2}{1+j}=\frac{-j+j8c+2C-j2C}{2C}\\ \Rightarrow Z_{in}=\frac{-j+j8c+2C-j2C}{2C}\\ Im\left(Z_{in}\right)=0\\ \Rightarrow C=\frac{1}{6}F=0.167F\approx 0.17F\end{array}$

$-\mathrm{V}+1\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}+4\left(\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}\right)-1\left(\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}\right)-1\left(\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}\right)=0$

$\Rightarrow 3\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-3\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=\mathrm{V}$

Also, for the second loop,

$4\left[\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}-\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}\right]+7\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}+\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}=0$

$\Rightarrow -3\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}+11\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=0$

From previous equation, we have

$3\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=3\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\mathrm{V}$

$11\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=11\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{11}{3}\mathrm{V}$

Substituting $11\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=11\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{11}{3}\mathrm{V}$ in $-3\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}+11\frac{\mathrm{d}{\mathrm{I}}_2}{\mathrm{d}\mathrm{t}}=0$ we have

$-3\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}+11\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{11}{3}\mathrm{V}=0$

$\Rightarrow 8\frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}-\frac{11}{3}\mathrm{V}=0$

$\Rightarrow \frac{\mathrm{d}{\mathrm{I}}_1}{\mathrm{d}\mathrm{t}}\left(\frac{24}{11}\right)=\mathrm{V}$

$\Rightarrow {\mathrm{L}}_{\mathrm{e}\mathrm{q}}=\frac{24}{11}=2.18\mathrm{H}$