An ac series circuit contains a resistor of 30 ohms, an inductor of 4.5 mH and a

Question

An ac series circuit contains a resistor of 30 ohms, an inductor of 4.5 mH and a variable capacitor. If the frequency of applied voltage is 400 Hz, to what setting should the capacitor be set if resonance is achieved? 43.2 mu F 156 mu F 35.18 mu F 133.2 mu F 7.09 mu F A multiple slit diffraction grating has a slit separation of 4 times 10-6 m. Find the wavelength of the monochromatic light that will have its 3 order bright fringe diffracted through an angle of 42 degree. (1 nm = 10^-9 m) A) 3.6 times 10^6 3.864 times 10^-6 0.8922 times 10^-6 2.1 times 10^-6 0.286 times 10^-6

Explanation / Answer

14. given, resistor, R = 30 ohms
   inductor, L = 4.5 mH = 4.5*10^-3 H
   frequency of applied voltage, f = 400 Hz
   let the capacitor be C
   then, for resonance
   1/wC = wL [ that is, capacitative reactance = inductive reactance]
   so, w^2 = 1/LC [ where w is angular frequency]
   so w = 2*pi*f
   hence
   4*pi^2*f^2 = 1/LC
   C = 1/4*pi^2*f^2*L = 1/4*pi^2*400^2 * 4.5*10^-3 = 3.518*10^-5 F = 35.18 micro F
   so the answer is c) 35.18 micro F
15. given, slit seperation, d = 4*10^-6 m
   order number, n = 3
   diffraction angle = 42 deg = 42*3.14/180 rad = 0.73266 rad
   now for multi slit diffraction
   the nth order bright fringe is found at angle theta
   wherer theta is given by
   theta = n*lambda/d
   where lambda is the wavelength of the light
   so,
   sin(0.73266) = 3*lambda/4*10^-6
   lambda = 8.9217*10^-7 m = 0.8922*10^-6 m
   so the answer is
   C) 0.8922*10^-6 m

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