Alternating Current Test

Result

Alternating Current Test - 54

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Figure shows a series LCR circuit connected to a variable frequency 230 V source.
The source frequency which drives the circuit in resonance is

A
4 Hz

B
5 Hz

C
6 Hz

D
8 Hz

Solution

Here,$$ L = 5H, C = 80\mu F = 80\times 10^{-6}F, R = 40\Omega $$
$$V_{rms}= 230 V$$

The resonant angular frequency is
$$\omega_r = \dfrac{1}{\sqrt{LC}} = \sqrt { \dfrac{1}{5\times 80\times 10^{-6}}} = 50 \, rad \, s^{-1}$$

$$\therefore \, \nu= \dfrac{\omega_r}{2\pi} = \dfrac{50}{2\pi} = 8 Hz$$
Question 2
1 / -0

A voltage of peak value 283 V and varying frequency is applied to series LCR combination in which R = 3$$\Omega$$, L = 25 mH and C = 400$$\mu$$F. Then the frequency (in Hz) of the source at which maximum power is dissipated in the above is

A
51.5

B
50.7

C
51.1

D
50.3

Solution

Here, $$V_0 \, = \, 283 \, V, \, R \, = \, 3\Omega, \, L \, = \, 25 \, \times \, 10^{-3} \, H$$
$$C \, = \, 400 \, \mu F \, = \, 4 \, \times \, 10^{-4} F$$
Maximum power is dissipated at resonance, for which

$$\nu \, = \, \dfrac{1}{2\pi \sqrt{LC}} \, = \, \dfrac{1 \, \times \, 7}{2 \, \times \, 22 \, \sqrt{25 \, \times \, 10^{-3} \, \times \, 4 \, \times \, 10^{-4}}}$$

$$= \, \dfrac{7 \, \times \, 10^3}{44\sqrt{10}} \, = \, 50.3 \, Hz$$
Question 3
1 / -0

A transmitter transmits at a wave length of 300 meters. A capacitor of a capacitance 9.6 $$\mu$$ F is being used. The value of the inductance for the resonant circuit is approximately

A
2.5 mH

B
2.5 $$\mu$$ H

C
2.5 nH

D
2.5 pH

Solution

V = f$$\lambda$$ where V is velocity and f is frequency
f = 1/ $$2 \pi \sqrt{LC}$$
$$L=\dfrac{({3 \times 10^2})^2}{4 \pi^2 \times 9.6 \times 10^{-6} \times ({3 \times 10^8)}^2}$$
$$L=2.5 \times 10^{-9}$$
0r $$L=2.5 nH$$
Question 4
1 / -0

A source of constant voltage $$V$$ is connected to a resistance $$R$$ and two ideal inductors $$L_{1}$$ and $$L_{2}$$ through a switch $$S$$ as shown. There is no mutual inductance between the two inductors. The switch $$S$$ is initially open. At $$t=0$$, the switch is closed and current beings to flow. Which of the following options is/are correct?

A
After a long time, the current through $$L_{1}$$ will be $$\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$$

B
After a long time, the current through $$L_{2}$$ will be $$\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$$

C
The ratio of the currents through $$L_{1}$$ and $$L_{2}$$ is fixed at all times $$(t\ >\ 0)$$

D
At $$t=0$$, the current through the resistance $$R$$ is $$\dfrac {V}{R}$$

Solution

After a long time current through $${ L }_{ 1 }$$ will be, $$\cfrac { V }{ R } \times \cfrac { { L }_{ 2 } }{ { L }_{ 1 }+{ L }_{ 2 } } $$
After a long time current through $${ L }_{ 2 }$$ will be ,$$\cfrac { V }{ R } \times \cfrac { { L }_{ 1 } }{ { R }_{ 2 } } $$
Time can be inferred from that $${ I }_{ L1 }:{ I }_{ L2 }={ L }_{ 2 }:{ L }_{ 1 }$$
The ratio of current is fixed all the time in $${ L }_{ 1 }$$ and $${ L }_{ 2 }$$.
Question 5
1 / -0

If R is resistance, L is inductance, C is capacitance, H is latent heat, and s is specific heat, then match the quantity given in Column I with the dimensions given in Column II.

Column I Column II
i. LC a. $$MA^{-2}L^2T^{-2}$$
ii. LR b. $$ML^2T^{-3}A^{-2}$$
iii. H c. $$T^2$$
iv. R d. $$M^2L^4T^{-5}A^{-4}$$

A
i. c, ii.b, iii.a, iv.d

B
i. a, ii.d, iii.a, iv.b

C
i. c, ii.d, iii.a, iv.b

D
i. c, ii.d, iii.a, iv.d

Solution

$$L = \dfrac{\phi}{L} = \dfrac{BA}{L} = \dfrac{FA}{q v L}= \dfrac{(MLT^{-2})(L^2)}{(AT)(LT^{-1}) (A)}$$
$$[L] = ML^2 T^{-2} A^{-2}$$
$$[R] = ML^2 T^{-3} A^{-2}$$
i) LC we know $$w = \dfrac{1}{\sqrt{LC}}$$
$$[LC] = \dfrac{1}{[w^2]} = \dfrac{1}{(T^{-1})^2} = T^2$$
ii) $$[LR] = ML^2 T^{-2} A^{-2} \times ML^2 A^{-2} T^{-3}$$
$$= M^2 L^4T^{-5} A^{-4}$$
iii) $$H = [L] = ML^2 T^{-2} A^{-2}$$
iv) $$R = ML^2 T^{-3} A^{-2}$$
Question 6
1 / -0

Directions For Questions

The capacitor of capacitance $$C$$ can be charged (with the help of a resistance $$R$$) by a voltage source $$V$$, by closing with switch $$S_{1}$$ while keeping switch $$S_{2}$$ open. The capacitor can be connected in series with an inductor 'L' by closing switch $$S_{2}$$ and opening $$S_{1}.$$

...view full instructions

After the capacitor gets fully charged, $$S_{1}$$ is opened and $$S_{2}$$ is closed so that the inductor is connected in series with the capacitor. Then,

A
at $$t=0$$, energy stored in the circuit is purely in the of magnetic energy

B
at any time $$t>0,$$ current in the circuit is in the same direction

C
at $$t>0,$$ there is no exchange of energy between the inductor and capacitor

D
at any time $$t>0$$, instantaneous current in the circuit may $$V\sqrt { \dfrac { C }{ L } } $$
Question 7
1 / -0

Two inductors of inductance L each are connected in series with opposite magnetic fluxes. The resultant inductance is (Ignore mutual inductance)

A
zero

B
L

C
2L

D
3L

Solution

When the two inductors are connected in series to one another, they are simply been added up and the direction of the magnetic flux does not have any affect on the net inductance of the two inductors.

$$L'=L_1+L_2$$

$$L'=L+L$$
$$L'=2L$$

So, option $$C$$ is correct.
Question 8
1 / -0

A $$100mH$$ inductor, a $$25\mu F$$ capacitor and a $$15\Omega$$ resistor are connected in series to a $$120V$$, $$50Hz$$ ac source. Calculate
(a) impedance of the circuit at resonance
(b) current at resonance
(c) resonant frequency

A
(a) $$16\Omega$$ (b) $$9A$$ (c) $$100Hz$$

B
(a) $$15\Omega$$ (b) $$8A$$ (c) $$100Hz$$

C
(a) $$15\Omega$$ (b) $$8A$$ (c) $$120Hz$$

D
(a) $$15\Omega$$ (b) $$7A$$ (c) $$100Hz$$

Solution

$$(a)$$ Reason occurs when $$X_C=X-L$$
$$\therefore z=15\Omega\\(b)I=\cfrac{V}{R}=\cfrac{120}{15}=8A\\(c)X_C=X_L\\\cfrac{1}{\omega_C}=\omega_L\\\omega^2=\cfrac{1}{LC}\\ \omega=\cfrac{1}{\sqrt{LC}}\\f=\cfrac{2\pi}{\sqrt{LC}}\times\cfrac{1}{2\pi}\\=100Hz$$
Question 9
1 / -0

A $$50W, 100V$$ lamp is to be connected to an AC mains of $$200V, 50Hz$$. What capacitance is essential to be put in series with the lamp?

A
9.2

B
8

C
10

D
12

Solution

Power$$=50w$$
$$V=100v\\I=\cfrac{P}{V}=.5A$$
The phase: potential across the capacitor must be lag by $$90° $$
$$(200)^2=100^2+V_2^2\\V_c=173.2V\\V_c=I_c\times x_e$$
$$x_e=346.4\Omega\\ X_c=\cfrac{1}{2\pi f_c}\\C=9.29 {\mu} f$$
Question 10
1 / -0

Find the maximum value of current when inductance of two henry is connected to $$150V,50$$ cycle supply

A
$$\cfrac { 3 }{ 2\sqrt { 2 } A } $$

B
$$\cfrac { 5 }{ 2\sqrt { 2 } A } $$

C
$$\cfrac { 6 }{ 2\sqrt { 2 } A } $$

D
$$\cfrac { 7 }{ 2\sqrt { 2 } A } $$

Solution

$$V_L=I\omega L\\I=\cfrac{150}{\sqrt2}\times\cfrac{1}{50\times2}\\ \cfrac{150}{2\sqrt2\times50}=\cfrac{3}{2\sqrt2}A$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Column I	Column II
i. LC	a. $$MA^{-2}L^2T^{-2}$$
ii. LR	b. $$ML^2T^{-3}A^{-2}$$
iii. H	c. $$T^2$$
iv. R	d. $$M^2L^4T^{-5}A^{-4}$$

Alternating Current Test - 54

Result

Alternating Current Test - 54

Score

-

Rank

-

SHARING IS CARING

Question 1

4 Hz

5 Hz

6 Hz

8 Hz

Solution

Question 2

51.5

50.7

51.1

50.3

Solution

Question 3

2.5 mH

2.5 $$\mu$$ H

2.5 nH

2.5 pH

Solution

Question 4

After a long time, the current through $$L_{1}$$ will be $$\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$$

After a long time, the current through $$L_{2}$$ will be $$\dfrac {V\ \ L_{2}}{R\ L_{1}+L_{2}}$$

The ratio of the currents through $$L_{1}$$ and $$L_{2}$$ is fixed at all times $$(t\ >\ 0)$$

At $$t=0$$, the current through the resistance $$R$$ is $$\dfrac {V}{R}$$

Solution

Question 5

i. c, ii.b, iii.a, iv.d

i. a, ii.d, iii.a, iv.b

i. c, ii.d, iii.a, iv.b

i. c, ii.d, iii.a, iv.d

Solution

Question 6

at $$t=0$$, energy stored in the circuit is purely in the of magnetic energy

at any time $$t>0,$$ current in the circuit is in the same direction

at $$t>0,$$ there is no exchange of energy between the inductor and capacitor

at any time $$t>0$$, instantaneous current in the circuit may $$V\sqrt { \dfrac { C }{ L } } $$

Question 7

zero

L

2L

3L

Solution

Question 8

(a) $$16\Omega$$ (b) $$9A$$ (c) $$100Hz$$

(a) $$15\Omega$$ (b) $$8A$$ (c) $$100Hz$$

(a) $$15\Omega$$ (b) $$8A$$ (c) $$120Hz$$

(a) $$15\Omega$$ (b) $$7A$$ (c) $$100Hz$$

Solution

Question 9

9.2

8

10

12

Solution

Question 10

$$\cfrac { 3 }{ 2\sqrt { 2 } A } $$

$$\cfrac { 5 }{ 2\sqrt { 2 } A } $$

$$\cfrac { 6 }{ 2\sqrt { 2 } A } $$

$$\cfrac { 7 }{ 2\sqrt { 2 } A } $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.