The voltage time (V-t) graph for triangular wave having peak value V0 is as shown in figure. The rms value of V in time interval from t=0 to T / 4 is

Question

The voltage time (V-t) graph for triangular wave having peak value V0 is as shown in figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e96f14e846d9b69b8343b5271c168c01-.png" /> The rms value of V in time interval from t=0 to T / 4 is (A) (V0/√3) (B) (V0/2) (C) (V0/√2) (D) None of these

Tardigrade Admin · Accepted Answer

V=(V0/T / 4) t ; V =(4 V0/T) t V rms =√ langle V2 rangle=(4 V0/T) √ langle t rangle =(4 V0/T) (∫ limits0T / 4 t2 d t/∫ limits01 / 4 d t) =(V0/√3)

Q. The voltage time $(V - t)$ graph for triangular wave having peak value $V_{0}$ is as shown in figure.

The rms value of $V$ in time interval from $t = 0$ to $T /4$ is

$\frac{V _{0}}{3}$

$\frac{V _{0}}{2}$

$\frac{V _{0}}{2}$

None of these

$V = \frac{V _{0}}{T /4} t; V = \frac{4 V _{0}}{T} t$
$V_{r m s} = ⟨ V^{2} ⟩ = \frac{4 V _{0}}{T} ⟨ t ⟩$
$= \frac{4 V _{0}}{T} ⎩ ⎨ ⎧ \frac{0 \int T /4 t ^{2} d t}{0 \int 1/4 d t} ⎭ ⎬ ⎫ = \frac{V _{0}}{3}$

Q. The voltage time (V−t) graph for triangular wave having peak value V0​ is as shown in figure. The rms value of V in time interval from t=0 to T/4 is

3​V0​​

2V0​​

2​V0​​

None of these

V=T/4V0​​t;V=T4V0​​t Vrms​=⟨V2⟩​=T4V0​​⟨t⟩​ =T4V0​​⎩⎨⎧​0∫1/4​dt0∫T/4​t2dt​⎭⎬⎫​=3​V0​​

Q. The voltage time $(V - t)$ graph for triangular wave having peak value $V_{0}$ is as shown in figure.

The rms value of $V$ in time interval from $t = 0$ to $T /4$ is

$\frac{V _{0}}{3}$

$\frac{V _{0}}{2}$

$\frac{V _{0}}{2}$

$V = \frac{V _{0}}{T /4} t; V = \frac{4 V _{0}}{T} t$
$V_{r m s} = ⟨ V^{2} ⟩ = \frac{4 V _{0}}{T} ⟨ t ⟩$
$= \frac{4 V _{0}}{T} ⎩ ⎨ ⎧ \frac{0 \int T /4 t ^{2} d t}{0 \int 1/4 d t} ⎭ ⎬ ⎫ = \frac{V _{0}}{3}$