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What is the applied voltage for a series RLC circuit when I_T = 3 mA, V_L = 30 V, V_C = 18 V, and R = 1000 ohms?

Answer: Option D

Explanation:

No answer description available for this question.

Lalit Garg said: (Mon, Nov 1, 2010 12:55:41 PM)

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

Swathi said: (Wed, May 4, 2011 11:54:30 AM)

XL=VL/I=10 and XC=VC/I=6 Z=sqrt of (100)2+(6-10)2=10016 i got V=IZ =310016 =300.239904077 i got but the given answer is not come how to solve to get that answer.

Vinodhine said: (Thu, May 19, 2011 11:36:50 AM)

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

Hussain said: (Wed, Jun 1, 2011 12:47:14 AM)

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: (Wed, Jun 15, 2011 05:22:30 AM)

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=4123.1056310^-3 V=12.369

Ankur said: (Sun, Dec 4, 2011 04:43:06 PM)

We have relation Vm= (VR^2+(VL-VC)^2)^1/2 VR= IR=310^-3 * 1000 = 3 volts VL= 30 volts VC= 18 volts therefore, Vm= 12.37

Stanley said: (Wed, Dec 14, 2011 12:15:07 PM)

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

Manikanta said: (Fri, Jan 27, 2012 06:07:09 AM)

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

Sukumar said: (Fri, Feb 24, 2012 09:35:24 PM)

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: (Thu, Mar 29, 2012 05:08:08 PM)

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: (Mon, May 21, 2012 04:43:01 PM)

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: (Wed, Jun 6, 2012 05:40:09 PM)

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

Nilesh said: (Sat, Jun 9, 2012 02:58:52 PM)

vs=squrt(squr(Itr)+squar(Vl-Vc)) =squrt(squr(31000*0.001)+squr(30-18)) =squrt(9+144) =12.37V

Anshuman said: (Thu, Aug 16, 2012 09:13:41 AM)

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: (Fri, Mar 22, 2013 10:30:02 AM)

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: (Sun, Apr 28, 2013 04:29:53 PM)

I Would like to give my solution to this problem which is as follow:- we know that voltage = currentImpedance < V = IZ >, 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 = CurrentImpedance, Mathematically "Voltage","V"; V = IZ. 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].