In a series L-C-R circuit the voltage across resistance , capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage

Monjur Alahi

In a series L-C-R circuit the voltage across resistance , capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be
A. 10/`sqrt 2` V
B. 10 V
C. 10`sqrt 2` V
D. 20 V


0 like 0 dislike
2 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A
`X_C` = `X_L` = R
V = `sqrt (V_R^(2) + (V_L - V_C)^(2))` but `V_L` = `V_C`
`therefore` V = `V_R` = 10
When capacitor is short circuited , `V_C` = 0
and `V_R` = `V_L`
`therefore` V = `sqrt V_(R)^(2) + V_(L)^(2)` or 10 = `sqrt 2V_L`
`thereforeV_L = 10/sqrt 2` V .

0 like 0 dislike
2 views
Talk Doctor Online in Bissoy App
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
In a series LR circuit, `X_L=R` and the power factor of the circuit is `P_1`.When a capacitor with capacitance C such that `X_C=X_L` is put in series

In a series LR circuit, power factor `(P_1)=R/Z` Here, Z =impedance = `sqrt(R^2+X_L^2)` Here, `Z=sqrt(R^2+R^2)=sqrt2R` `thereforeP_1=1/sqrt2` In a series LCR circuit power factor `(P_2)=R/Z` where,`Z=sqrt(R^2+(X_L-X_C)^2)=R` `thereforeP_2=1` Hence,`P_1/P_2=1/sqrt2`

2 Answers 1 views
A resistance and inductance are connected in series across a voltage, `V=283sin314t` The current is found to be `4sin(314t-pi//4)`. Find the value of

In L-R series cirucit current lags the voltage by an angle, `phi = tan^(-1)(X_(L)/R)` Here, `phi = pi/4` `X_(L) = R` or `omegaL=R` `[omega=314 rad s^(-1)` `therefore` 314 L =...

2 Answers 1 views
Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values `1000 Omega, 1muF` an

The rms value of voltage across the source, `V_(rms) = V_(0)/sqrt(2)` comparing the given equation with general equation, `V_(0) = 100sqrt(2),omega=1000rads^(-1)` Also, R=`1000Omega`, C=`1muF= 1xx10^(-6)`F L=2H `V_(rms)=(100sqrt(2))/(sqrt(2)) = 100V` `therefore...

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
In R-L-C series circuit, the potential difference across each element is 20 V. Now the value of hte resistance alone is doubled, then PD across R, L a

Correct Answer - A Given, the potential difference across each element is same i.e., 20V, so the circuit is at resonance and we have `V=V_(R)=20V` If the value of R is...

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
In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the change in inductance, so that the circuit remains in

Correct Answer - A At resonance `omegaL=1/(omegaC) rArr l prop 1/C`.

2 Answers 1 views
A 2000 Hz, 20 volt source is connected to a resistance of 20 ohm, an inductance of `0.125//pi` H and a capacitance of `500/pi` nF all in series. Calcu

Correct Answer - 50 power`(P)=I_(rms)^(2)R , P=[20/sqrt(R^(2)+(omegaL-1/(omegaC))^(2))]^(2)xxR` where `omega=2xxpixx2000` `DeltaH=(ms)Deltatheta=P(Deltat) rArrDeltat=((ms)Deltatheta)/P=(100xx10)/P=50 sec. rArrx=8`

2 Answers 1 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