A generator with an adjustable frequency of oscillation is wired in series to an inductor of L=2.50 mH and a capacitor of C=3.00 uF. At what frequency does the generator pro-duce the largest possible current amplitude in the circuit? (Answer: 1.84 kHz)

Question

anonymous · Answer

At resonance frequency, the generator will produce the largest possible current amplitude in the circuit 
solution : wL = wC
              w = 1 divided by sqrt(LC)
       frequency (v) = 1 divided by 2*pi* sqrt(LC)
     on putting the values i m getting an answer 1.79 kHz

anonymous · Answer

base on the question, 

we know 
L=2.50 mH = 2.5 x 10^-3 H 
C=3.00 uF = 3 x 10 ^-6 F 

XL = XC 
w*L = 1/(wC)
w^2*L*C = 1 
w^2 = 1/LC 
w = sqrt(1/LC) 
w = sqrt(1/(2.5 x 10^-3 x 3 x 10^-6)) 
w = sqrt(1/(7.5x10^-9)) 
w = sqrt(1.3x10^8) 
w = 11547 

w = 2*pi*f 

f = w/(2*pi) 
f =11547/(2*3.14) 
f = 11547/6.28 
f = 1838 Hz 
f = 1.84 KHz

@Idealist right???

anonymous · Answer

it was my horrible mistake 
the formula was wL = 1 divided by wC 
i had applied the right formula but out of curiosity i wrote wL=wC

anonymous · Answer

no problem @niksva ur right too.,  :)