Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Alternating Current question

2022 · 29 Jun · Shift 2 · Q70
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Alternating Current
  5. /2022 · 29 Jun · Shift 2 · Q70

Alternating Current question

2022 · 29 Jun · Shift 2 · Q70

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
An inductor of 0.5 mH, a capacitor of 200 μ\muμ F and a resistor of 2 Ω\OmegaΩ are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ‾\underline{\hspace{2cm}}​×\times× 102 Hz.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Since the current is in phase with the emf in a series RLCRLCRLC circuit, the circuit is at resonance.

  2. For resonance in a series RLCRLCRLC circuit,

XL=XCX_L = X_CXL​=XC​

which gives

ωL=1ωC\omega L = \frac{1}{\omega C}ωL=ωC1​

So,

ω2=1LC\omega^2 = \frac{1}{LC}ω2=LC1​

and hence

f=12πLCf = \frac{1}{2\pi\sqrt{LC}}f=2πLC​1​
  1. Given:
L=0.5 mH=0.5×10−3 HL = 0.5\,\text{mH} = 0.5 \times 10^{-3}\,\text{H}L=0.5mH=0.5×10−3H C=200 μF=200×10−6 FC = 200\,\mu\text{F} = 200 \times 10^{-6}\,\text{F}C=200μF=200×10−6F
  1. Compute LCLCLC:
LC=(0.5×10−3)(200×10−6)LC = (0.5 \times 10^{-3})(200 \times 10^{-6})LC=(0.5×10−3)(200×10−6) =100×10−9=10−7= 100 \times 10^{-9} = 10^{-7}=100×10−9=10−7
  1. Then
LC=10−7=10−3.5\sqrt{LC} = \sqrt{10^{-7}} = 10^{-3.5}LC​=10−7​=10−3.5

So,

f=12π×10−3.5f = \frac{1}{2\pi \times 10^{-3.5}}f=2π×10−3.51​

Using 10−3.5≈3.162×10−410^{-3.5} \approx 3.162 \times 10^{-4}10−3.5≈3.162×10−4,

2πLC≈6.283×3.162×10−42\pi\sqrt{LC} \approx 6.283 \times 3.162 \times 10^{-4}2πLC​≈6.283×3.162×10−4 ≈1.986×10−3\approx 1.986 \times 10^{-3}≈1.986×10−3

Thus,

f≈11.986×10−3≈503.5 Hzf \approx \frac{1}{1.986 \times 10^{-3}} \approx 503.5\,\text{Hz}f≈1.986×10−31​≈503.5Hz
  1. The question asks for frequency in the form
‾×102 Hz\underline{\hspace{2cm}} \times 10^2\,\text{Hz}​×102Hz

Therefore,

503.5 Hz≈5.0×102 Hz503.5\,\text{Hz} \approx 5.0 \times 10^2\,\text{Hz}503.5Hz≈5.0×102Hz

So the required integer is

5\boxed{5}5​
PreviousNext

More from Alternating Current

  • In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be : (f1 = original resonant frequency, Q1 = original quality factor)2022 · MCQ
  • A series LCR circuit with R=11250​Ω and XL​=1170​Ω is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be ​…2022 · Numerical
  • For the given circuit the current i through the battery when the key in closed and the steady state has been reached is ​. Includes diagram2021 · MCQ
  • An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to : Includes diagram2021 · MCQ
  • A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8 Ω, L = 24 mH and C = 60 μ F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ​…2021 · Numerical
  • For the given circuit, comment on the type of transformer used. Includes diagram2021 · MCQ
  • An AC current is given by I = I1 sin ω t + I2 cos ω t. A hot wire ammeter will give a reading :2021 · MCQ
  • Match List - I with List - II Choose the most appropriate answer from the options given below : Includes table2021 · MCQ