Question 1 |
Figure Shows the waveform of the current passing through an inductor of
resistance 1\Omega and inductance 2 H. The energy absorbed by the inductor in the first four seconds is 
144J | |
98J | |
132J | |
168J |
Question 1 Explanation:
For 0 \lt t \lt 2s current varies linearly with time and given as i(t)=3t and for 2s \lt t \lt 4s current is constant, i(t)=6A.
The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,
E_L=\int_{0}^{T}Li\frac{di}{dt}=E_{L_1}+E_{L_2}
E_{L_1}=\int_{0}^{2}Li \left (\frac{di}{dt} \right )dt
\;\;=\int_{0}^{2} 2 \times 3t \times 3\; dt
\;\;=18\int_{0}^{2}t\; dt
\;\;=18 \times \left.\begin{matrix} \frac{t^2}{2} \end{matrix}\right|^2_0
\;\;= 18 \times \left [ \frac{4}{2}-0 \right ]=36J
The energy absorbed by the inductor in (2\rightarrow 4) second
E_{L_2}=\int_{2}^{4}Li\left ( \frac{di}{dt}dt \right )
\;\;=\frac{2}{4}2 \times 6 \times 0 \;dt=0 J
A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.
The energy dissipated by the resistance in 4 sec.
E_R=\int_{0}^{T}i^2 R\;dt
\;\;=\int_{0}^{2}(3t)^2 \times 1 \;dt+\int_{2}^{4}6^2 \times 1 \; dt
\;\;=9\int_{0}^{2}t^2\;dt +36\int_{2}^{4}1 \; dt
\;\;=9 \times \left.\begin{matrix} \frac{t^3}{3} \end{matrix}\right|_0^2+36t|_2^4
\;\;=9 \times \left ( \frac{8}{3} \right )+36 \times 2
\;\;=24+72=96J
The total energy absorbed by the inductor in 4 sec
=96 J +36 J =132 J
The energy absorbed by the inductor (Resistance neglected) in the first 2 sec,
E_L=\int_{0}^{T}Li\frac{di}{dt}=E_{L_1}+E_{L_2}
E_{L_1}=\int_{0}^{2}Li \left (\frac{di}{dt} \right )dt
\;\;=\int_{0}^{2} 2 \times 3t \times 3\; dt
\;\;=18\int_{0}^{2}t\; dt
\;\;=18 \times \left.\begin{matrix} \frac{t^2}{2} \end{matrix}\right|^2_0
\;\;= 18 \times \left [ \frac{4}{2}-0 \right ]=36J
The energy absorbed by the inductor in (2\rightarrow 4) second
E_{L_2}=\int_{2}^{4}Li\left ( \frac{di}{dt}dt \right )
\;\;=\frac{2}{4}2 \times 6 \times 0 \;dt=0 J
A pure indictor does not dissipate enegy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy stored in the inductor and the energy dissipated in the resistor.
The energy dissipated by the resistance in 4 sec.
E_R=\int_{0}^{T}i^2 R\;dt
\;\;=\int_{0}^{2}(3t)^2 \times 1 \;dt+\int_{2}^{4}6^2 \times 1 \; dt
\;\;=9\int_{0}^{2}t^2\;dt +36\int_{2}^{4}1 \; dt
\;\;=9 \times \left.\begin{matrix} \frac{t^3}{3} \end{matrix}\right|_0^2+36t|_2^4
\;\;=9 \times \left ( \frac{8}{3} \right )+36 \times 2
\;\;=24+72=96J
The total energy absorbed by the inductor in 4 sec
=96 J +36 J =132 J
Question 2 |
A segment of a circuit is shown in figure. V_{R} = 5V, V_{C} = 4 \sin 2t. The voltage V_{L} is
given by
3-8 \cos 2t | |
32 \sin 2t | |
16 \sin 2t | |
16 \cos 2t |
Question 2 Explanation:
By KCL,
I_P+I_Q+I_C+I_L=0
2+2+I_C+I_L=0
But, I_C=C \times \frac{dv}{dt}
\;\; =1 \times \frac{d}{dt}(4 \sin 2t)
\;\;=(8 \cos 2t)
\therefore \;\; I_L=-(2+1+8 \cos 2t)
\;\; \;\;=-3-8 \cos 2t
\therefore \;\; V_L=L\left ( \frac{di}{dt} \right )
\;\;\;=2 \times 2 \times 8 \sin 2t
\;\;\;=32 \sin 2t
NOTE: KCL is based on the law of conservation of charges.
I_P+I_Q+I_C+I_L=0
2+2+I_C+I_L=0
But, I_C=C \times \frac{dv}{dt}
\;\; =1 \times \frac{d}{dt}(4 \sin 2t)
\;\;=(8 \cos 2t)
\therefore \;\; I_L=-(2+1+8 \cos 2t)
\;\; \;\;=-3-8 \cos 2t
\therefore \;\; V_L=L\left ( \frac{di}{dt} \right )
\;\;\;=2 \times 2 \times 8 \sin 2t
\;\;\;=32 \sin 2t
NOTE: KCL is based on the law of conservation of charges.
Question 3 |
In the figure,
Z_{1}=10\angle -60 ^{\circ} , Z_{2}=10\angle 60^{\circ} , Z_{3}=50\angle 53.13^{\circ}.
The venin impedance seen form X-Y is
Z_{1}=10\angle -60 ^{\circ} , Z_{2}=10\angle 60^{\circ} , Z_{3}=50\angle 53.13^{\circ}.
The venin impedance seen form X-Y is
56.66\angle 45^{\circ} | |
60\angle 30^{\circ} | |
70\angle 30^{\circ} | |
34.4\angle 65^{\circ} |
Question 3 Explanation:
By Thevenin's theorem
Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3
\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3
\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13
\;\;=56.66\angle 45^{\circ}\Omega
Z_{Th}=Z_{X-Y}=Z_1||Z_2+Z_3
\;\;=\frac{Z_1 \times Z_2}{Z_1+Z_2}+Z_3
\;\;=\frac{10\angle -60 \times 10\angle 60}{10\angle -60 + 10\angle 60}+50\angle 53.13
\;\;=56.66\angle 45^{\circ}\Omega
Question 4 |
Two conductors are carrying forward and return current of +I and -I as
shown in figure. The magnetic field intensity H at point P is
\frac{I}{\pi d}\vec{Y} | |
\frac{I}{\pi d}\vec{X} | |
\frac{I}{2 \pi d}\vec{Y} | |
\frac{I}{2 \pi d}\vec{X} |
Question 4 Explanation:
\bar{H}=\bar{H_1}+\bar{H_2}=\frac{I}{2 \pi d}\vec{y}+\frac{I}{2 \pi d}\vec{y}=\frac{I}{ \pi d}\vec{y}
Question 5 |
Two infinite strips of width w m in x -direction as shown in figure, are carrying
forward and return currents of +I and -I in the z - direction. The strips
are separated by distance of x m. The inductance per unit length of the
configuration is measured to be L H/m. If the distance of separation between
the strips in snow reduced to x/2 m, the inductance per unit length of the
configuration is
2L H/m | |
L/4 H/m | |
L/2 H/m | |
4L H/m |
Question 5 Explanation:
Inductance is proportional to separation between the strips so when separation is reduced to x/2 then inductance will become L/2 H/m.
There are 5 questions to complete.