Question

he amplitude, or magnitude, of a sinusoidal source is the maximum value of the source. What is the amplitude of the voltage source described asv(t)=50cos(2000t−45∘) mV?

Answer 1

Answer:

Amplitude is $5\times 10^{-2} \ V.$

Explanation:

We need to find the amplitude of voltage source :

$V(t)=50\times cos(2000t-45^o)\ mV=5\times 10^{-2}\times cos(2000t-45^o)\ V.$

Amplitude of any function is its maximum value from its equilibrium position.

Therefore, for amplitude to be maximum term containing cos should be greatest . and we know cos $\theta$ has maximum value = 1.

Therefore, amplitude , $A = 5\times 10^{-2} \ V.$

Hence, this is the required solution.

Answer 2

Answer:

Explanation:

V = 50 cos (2000t - 45) mV

it is the time varying voltage.

by comparison with the standard equation

v = Vo Cos (ωt - Ф)

where, Vo is the amplitude of the voltage that means it is the maximum value of the voltage.

So, Vo = 50 mV.