1-T-t-1-t-2-Vsin-t-V-dt-t-1-and-t-2-are-solution-to-Vsin-t-V-V-V-V-0-and-t-1-lt-t-2-

Question Number 737 by 123456 last updated on 08/Mar/15

\frac{1}{T} \underset{t_{1}}{\int^{t_{2}}} V \sin ω t - V_{γ} d t = ?

t_{1} a n d t_{2} a r e s o l u t i o n t o

V \sin ω t = V_{γ}

V ⩾ V_{γ}

V_{γ} ⩾ 0

a n d t_{1} < t_{2}

Commented by prakash jain last updated on 08/Mar/15

V \sin ω t = V_{γ}

t_{2} = t_{1} + 2 π k T ∵ ω = \frac{2 π}{T}

Answered by prakash jain last updated on 08/Mar/15

\frac{1}{T} \int V \sin ω t - V_{γ} d t

= \frac{1}{T} {[- \cos ω t - V_{γ} t]}_{t_{1}}^{t_{2}} = \frac{V_{γ}}{T} [t_{1} - t_{2}] = \frac{V_{γ}}{T} (- \frac{2 π k}{T})

Since \sin {w t}_{1} = \sin {w t}_{2} \Rightarrow \cos ω t_{1} = \cos {w t}_{2}

The given integral = - \frac{2 π k}{T^{2}} V_{γ}, k > 0

V is assumed to be constant amplitude .