When an AC voltage, of variable frequency is applied to series L-C-R circuit , the current in the circuit is the same at 4 kHz and 9 kHz. The current

Monjur Alahi

When an AC voltage, of variable frequency is applied to series L-C-R circuit , the current in the circuit is the same at 4 kHz and 9 kHz. The current in the circuit is maximum at
A. 5 kHz
B. 6.5 kHz
C. 4.2 kHz
D. 6 kHz


0 like 0 dislike
3 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - D
(d) `I_(1) = I_(2)`
`therefore omega_(1)L - 1/(omega_(1)C) = 1/(omega_(2)C) - omega_(2)L`
Solving this, we get
`1/sqrt (LC)`= responance frequency = `sqrt (omega_(1)omega_(2))`
At resonance frequency , current will be maximum .

0 like 0 dislike
3 views
Talk Doctor Online in Bissoy App
A series LCR circuit with R=20`Omega` , L=1.5 H and C=35`muF` is connected to a variable frequency 200 V ac supply. When the frequency of the supply e

When natural frequency and supply frequency and supply frequency are equal, resonance occurs. `thereforeX_L=X_CthereforeZ=R` `P=V^2/Z=V^2/R=(200times200)/20=2000 W`

2 Answers 1 views
A series LCR circuit is connected to a variable frequency 230 V source , L=5.0 H, `C=80muF,R=40Omega` Obtain impedance of the circuit and aplitude of

At resonance impedance, `Z=R=40Omega` `thereforeI_(rms)=V_(rms)/R=(sqrt2times230)/40=8.1 A`

2 Answers 1 views
State the condition under which the phenomenon of resonance occurs in series LCR circuit when ac voltage is applied. In a series LCR circuit, the curr

Condition of resonance in a series LCR circuit: `omegaL=1/(omegaC)` where `omega`= angular frequency of the source, L=self inductance of the coil and C=capacitance. Accordingly to the question, electric current and...

2 Answers 2 views
When the frequency of the ac voltage applied to a series LCR circuit is gradually increased from a low value , the impedance of the circuit.

Correct Answer - C We Know`Z=sqrt((omegaL-1/(omegaC))^2+R^2)` IF , `(dZ)/(domega)=0`then `omega=1/sqrt(LC)` when `omegalt1/sqrt(LC)` then `(dZ)/(domega)le0` i.e., Z is a decreasing function. Again, when `omegagt1/sqrt(LC)`then`(dZ)/(domega)gt0` i.e., Z is a increasing function.

2 Answers 1 views
A sinusoidal voltage of frequency 60 Hz and peak value 150 V is applied to a series L-R circuit, where `R=20Omega` and L=40 mH. a) Compute T, `omega,

`As I_(rms) = V_(rms)/X_(L)` Where, `X_(L)=omegaL=2pifL` is the reactance of the inductor Here, f=50H, L=44mH =`44xx10^(3)`H, `V_(rms)=220V` `therefore X_(L)=2pi xx 50 xx 44 xx 10^(-3)= 13.82Omega` `therefore` rms value...

2 Answers 1 views
A series L-C-R circuit with R = `20 Omega`, L = 1.5 H and C =` 35muF` is connected to a variable frequency 200 V, AC supply. When the frequency of the

When the frequency of the supply equals the natural frequeny of the circuit, the circuit is said to be in resonance. At resonance, ? Z=R = `20Omega` Since, the L-C-R circuit...

2 Answers 1 views
An sinusoidal voltage of peak value 300 V and an argular frequency `omega` = 400 `rads^(-1) ` is applied to series L-C-R circuit , in which R = ` 3 Om

Correct Answer - B The impedance of the circuit is Z = `sqrt R^(2) + (underset(L)(X)- underset(C )(X))^(2)` `underset(L)(X)= omegaL = 400 xx 20 xx 10^(-3)= 8 H` `X_(C)` = `1omegaC...

2 Answers 1 views
Assertion: In series, L-C-R voltage across capacitor is always less than the applied voltage. Reason: In series L-C-R circuit, `V = sqrt((V^(2) + (V_(

Correct Answer - D d) `V_(R) larrV`, but `V_(C)` or `V_(L)` can be greater than V also.

2 Answers 1 views
A series circuit consists of a resistance inductance and capacitance.The applied voltage and the current at any instant are given by `E=141.4 cos(5000

Correct Answer - `4 muF,R=141.4/5 Omega` Here phase difference `(0)=360^(0)` & `omega=5000` at Resonance `C=1/(5000)^(2)xx1/(0.01)` `R=V_(0)/I_(0)=141.4/5`

2 Answers 4 views
An alternating voltage of frequency `f` is applied across a series `LCR` circuit.Le `f_(r)` be the resonance frequency for the circuit.Will the curren

`X_(L)=2pifL` and `X_(C)=1/(2pifC)` (i)When `f gt f_(r),X_(L)` is large and `X_(C)` is small.The circuit is induction current lags behind the applied voltage. (ii)When `fltf_(r),X_(L)` is small and `X_(C)` is large.The...

2 Answers 2 views
X