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? A) 43.2 mu F B) 156 mu F C) 35.18 mu F D) 133.2 mu F E) 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 B) 3.864 times 10^-6 C) 0.8922 times 10^-6 D) 2.1 times 10^-6 E) 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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