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Standard XII

Physics

Impedance Triangle

The input sig...

Question

The input signal given to a CE amplifier having a 15t+ . The voltage gain of 150 is Vi = 2 cos corresponding output signal will be - 15t+ (1) 300 cos 15t+ (2) 300 cos 15t+ (3) 75 cos 57th

300 cos (15 t + \frac{4 π}{3})

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100 cos (15 t + \frac{4 π}{3})

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30 cos (15 t + \frac{4 π}{3})

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300 sin (15 t + \frac{4 π}{3})

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Solution

The correct option is A $300 cos (15 t + \frac{4 π}{3})$

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Similar questions

Q. The input signal given to a a CE amplifier having a voltage gain of

150

V_{i} = 2 cos (15 t + \frac{π}{3})

. The corresponding output signal will be

Q. The input signal given to a a CE amplifier having a voltage gain of

150

V_{i} = 2 cos (15 t + \frac{π}{3})

. The corresponding output signal will be

Q. The input signal given to a a CE amplifier having a voltage gain of

150

V_{i} = 2 cos (15 t + \frac{π}{3})

. The corresponding output signal will be

Q. Assertion :consider

Δ

ABC whose verticies are A = (3 + 4 cos

Θ, - 5 + 4 s i n Θ

)

B = (3 + 4 c o s (Θ + \frac{2 π}{3}), - 5 + 4 s i n (Θ + \frac{2 π}{3}))

C = (3 + 4 c o s (Θ + \frac{4 π}{3}), - 5 + 4 s i n (Θ + \frac{4 π}{3}))

The orthocentre of the triangle is (3, 5) Reason: consider

Δ

ABC whose verticies are A = (3 + 4 cos

Θ, - 5 + 4 s i n Θ

)

B = (3 + 4 c o s (Θ + \frac{2 π}{3}), - 5 + 4 s i n (Θ + \frac{2 π}{3}))

C = (3 + 4 c o s (Θ + \frac{4 π}{3}), - 5 + 4 s i n (Θ + \frac{4 π}{3}))

The triangle ABC is equilateral

$I f x c o s θ = y c o s (θ + \frac{2 π}{3}) = z c o s (θ + \frac{4 π}{3}), t h e n t h e v a l u e o f \frac{1}{x} + \frac{1}{y} + \frac{1}{z} i s e q u a l t o$

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The input signal given to a CE amplifier having a 15t+ . The voltage gain of 150 is Vi = 2 cos corresponding output signal will be - 15t+ (1) 300 cos 15t+ (2) 300 cos 15t+ (3) 75 cos 57th

The correct option is A 300cos(15t+4π3)

The correct option is A $300 cos (15 t + \frac{4 π}{3})$