### Discussion :: RLC Circuits and Resonance - General Questions (Q.No.1)

Lalit Garg said: (Nov 1, 2010) | |

VL=I*XL , VC=I*XC XL=?,XC=?,R=1000 Z=UNDERROOT OF R2+(XC-XL)2 V=I*Z=12.37 |

Swathi said: (May 4, 2011) | |

XL=VL/I=10 and XC=VC/I=6 Z=sqrt of (100)2+(6-10)2=10016 i got V=I*Z =3*10016 =300.239904077 i got but the given answer is not come how to solve to get that answer. |

Vinodhine said: (May 19, 2011) | |

Swathi, you have to consider the R value as 1 Kohm, and dont forget to note the current is in milliampere so resistance will be in Kiloohms |

Siva said: (May 25, 2011) | |

Vinodhine is correct. |

Hussain said: (Jun 1, 2011) | |

Vinodhine the value of R = 1000 that is 1K but u have taken as 100. Galit garg is right. I = v/z => v = I * z = 3m * sqrt(17)*k = 3 * 4.123 = 12.37v |

Amulya said: (Jun 15, 2011) | |

If we do them as if they are easy they will be easy...n if we think them to be difficult they will be difficult.....ALL THE BEST FRNDS...!!!!!!!!! V=IZ z=impedence Z=(R^2+(XL-XC)^2)^1/2 XL=VL/I=10000 XC=VC/I=6000 ----> Z=4123.1056 V=4123.1056*3*10^-3 V=12.369 |

Sarojini said: (Dec 2, 2011) | |

Hussain is correct. |

Ankur said: (Dec 4, 2011) | |

We have relation Vm= (VR^2+(VL-VC)^2)^1/2 VR= I*R=3*10^-3 * 1000 = 3 volts VL= 30 volts VC= 18 volts therefore, Vm= 12.37 |

Stanley said: (Dec 14, 2011) | |

Lets train ourselves to use SI units always, by so doing you will not mess. Thanks. |

Manikanta said: (Jan 27, 2012) | |

Why can't we write kvl for this but I'm not getting answer. |

Yugandhar said: (Jan 31, 2012) | |

VR=3v; vl=30v; vc=18v; vi=(3^2+(30-18)^2)^1/2; |

Sukumar said: (Feb 24, 2012) | |

According to Lalit Garg Givendata:- Vl=30;Vc=18;It=0.03A; Xl=Vl/It--->30/0.03=1000 Xc=Vc/It--->18/0.03=600 Z=(R^2+(Xc-Xl)^2)^1/2 then z=1077.03 Vsource=Z*It=32.31V but this is not valid with options. |

Biru said: (Mar 29, 2012) | |

Given voltage across inductor = 30v and capacitor=18v. R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt. So applied voltage = 30V + 18V + 3V = 51V. |

Saikrishna said: (May 21, 2012) | |

Given voltage across inductor = 30v and capacitor=18v. R= 1000 ohms, so voltage across "R" is 1000*3mA = 3volt. So applied voltage = 30V + 18V + 3V = 51V. Because as the answer says 12.37volts. If source is such less how can the drop of 30v occurs at inductor? |

Badar said: (Jun 6, 2012) | |

Because votlage across inductor and capacitor are opposite in phase about 180 degree. They cancel the effect of each other :-). |

Nilesh said: (Jun 9, 2012) | |

vs=squrt(squr(It*r)+squar(Vl-Vc)) =squrt(squr(3*1000*0.001)+squr(30-18)) =squrt(9+144) =12.37V |

Anshuman said: (Aug 16, 2012) | |

Vl=30;Vc=18;It=0.03A; Xl=Vl/It--->30/0.03=1000 Xc=Vc/It--->18/0.03=600 Z=(R^2+(Xc-Xl)^2)^1/2 then z=1077.03 Vsource=Z*It=32.31V |

Prakash said: (Mar 22, 2013) | |

We have to find impedance because we have been asked to find out AC voltage source. Impedance is the resistance to AC voltage, thats why we take out impedance (Z) with the help of reactance of both L and C. Rest all the calculation have been explained by others. |

Khalid Anis said: (Apr 28, 2013) | |

I Would like to give my solution to this problem which is as follow:- we know that voltage = current*Impedance < V = I*Z >, We know that Impedance= [Square Root of {R^2 +Square of ((Inductive resistance)-(Capacitive reactance)). Mathematically "Impedance","Z" can be written as Z = [{R^2+(XL-XC)^2}]^2, We know that Voltage = Current*Impedance, Mathematically "Voltage","V"; V = I*Z. Now I will come to the solution:Z = [(1000^2+(10000-60000^2]^1/2, V = (4123.105626)(3*10^-3), V = 12.36931688. Therefore answer to the solution is V = 12.37, [b]. |

Rajkamal Gupta said: (Jun 24, 2013) | |

From phasor diagram it is clear that VL > Vc V = sqrt[Vr^2+(VL-Vc)^2]. On putting the given value on formula as given unit we get exact answer. |

Surya said: (Sep 7, 2013) | |

Can anyone please difference between impedance and resistance. Can we write v = i*z (z=impedance ). |

Govind Singh Chouhan said: (Nov 25, 2013) | |

Given data: R = 1000R. Vc = 18V. VL = 30V. I = 3mA. We know that, Z=sqrt((R ^{2})+(XL-XC)^{2})And XL = VL/I ==> 30/0.003 = 10K. And XC = VC/I ==> 18/0.003 = 6K. So know put the value of XL and XC in in formula of Z so we can say that: Z = sqrt(1000 ^{2}+(10K-6K)^{2})Z = sqrt(1000000+16000000) Z = sqrt(17000000) Z = 4123 or Z = 4.123K. Now it applied voltage to the circuit is: V=I*Z V=0.003*4123 V=12.369 or 12.37V. So applied voltage to this circuit is 12.37V. Thanks. |

Swapnil said: (Apr 19, 2014) | |

Vr = 1000*10^-3 = 3v. VL = 30v. vc = 18v. V = square root of(3^2+(30-18)^2). = square root of (9+144). = square root of (153). = 12.3691. Answer = B. |

Vinit said: (May 24, 2015) | |

E = I*R. E = 0.003*1000. E = 3 volts. The reactive volts are: VX = VL - VC. VX = 30 - 18. VX = 12 volts. The applied volts is: Volts = Sqrt(VR^2 + VX^2). Volts = Sqrt (3^2 + 12^2). Volts = Sqrt(153). Volts = 12.3693. |

Anil Singh said: (Sep 4, 2015) | |

Vr = IxR = 0.003 Ampx1000 Ohms = 3 Volts. Now we have VL = 30V, Vc = 18V and Vr = 3V. Applied voltage is given by. Vs = SQRT of [Vr^2 + (VL-Vc) ^2]. = [3^2 + (30-18) ^2] = [9 + (12^2) ] = [9 + 144] = 153. Therefore Vs = Square root of 153. Vs = 12.369 Volts or Say 12.37 Volts here is the answer. |

Mahu said: (Sep 10, 2015) | |

By tricky its sure answer will be A or B but why B tell me please. |

Leelavathi said: (Mar 2, 2016) | |

Firstly the formula is v = IZ. Xl = vl/I = 30/3*10^-3 = 10,000. Xc = vc/I = 18/3*10^-3 = 6000. Z = sqr of (R^2 + (xl-xc)). Z = sqr of (1000^2 + (10000-6000)). Z = sqr of (1004000) = 1001.998004. Now, V = 3 * 10^-3 * 1001.998004 = 3.005994. Approximately ans = 3.00 v. I don't know whether my answer is right or wrong. If I am wrong, please tell me my mistake. |

Khan said: (Sep 16, 2016) | |

VL = 30 <90, VC =1 8<-9 0, VR = 3. hence V = sqrt(12^2 + 3^2) = 12.37. |

Weerasinghege Maduka said: (Oct 23, 2016) | |

Create voltare trangle. v^2 = (VL-VC)^2 + VR^2 So, VR= I * R VR= 3mA*1000Ohms VR=3V V^2=(XL-XC)^2+VR^2, =(30-18)^2+3^2, =12^2 + 9, =144 + 9, => V^2 = 153. V = 12.36. |

Ch Manikantha said: (Nov 30, 2016) | |

Vr = i * r =3mA * 1k = 3v Vs = √(Vr^2 + (Vc - Vl)^2). Vs = √(3^2 + (30 - 18)^2) = 12.36. |

Mukesh said: (May 31, 2017) | |

Well said, Thanks, @Govind Singh Chouhan. |

Kailas Jagtap said: (Dec 5, 2017) | |

V=IR=3*10^-3*1000=3V, 3^2+12^2=151, = 12.37. |

Yasir Shaikh said: (Apr 24, 2018) | |

The voltage drop across the resistor is E = I * R. E = 0.003 * 1000, E = 3 volts. The reactive volts are; VX = VL - VC. VX = 30 - 18. VX = 12 volts. The applied volts is; Volts = √(VR^2 + VX^2), Volts = √(3^2 + 12^2), Volts = √(153), Volts = 12.3693. |

G.Satish Kumar said: (May 28, 2018) | |

V^2 = Vr^2+(VL-VC)^2. V^2 = 3^2+(30-18)^2, =9+144, V^2= 153. V = 12.369. |

Bilal Niazi said: (Aug 23, 2019) | |

Thanks @ Singh Chouhan. |

Bilal Niazi said: (Aug 23, 2019) | |

Thanks @Govind Singh Chouhan. |

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