Question

The voltage supplied to a circuit is given by $V = V_0t^{3/2}$, where t is time in second. Find the RMS value of voltage for the period t = 0 to t = 1s :-

• A. $V_0/2$
• B. $V_0$
• C. $3V_0/2$
• D. $2V_0$

Answer 1

Answer: (A)

$V_0/2$

Answer 2

The RMS value of a voltage $V(t)$ over a time interval $[0, T]$ is given by: $V_{RMS} = \sqrt{\frac{1}{T} \int_{0}^{T} [V(t)]^2 dt}$

Given $V(t) = V_0t^{3/2}$ and the time interval is from $t=0$ to $t=1$ s, so $T=1$.

1. Square the voltage function: $[V(t)]^2 = (V_0t^{3/2})^2 = V_0^2 t^{(3/2) \times 2} = V_0^2 t^3$

2. Integrate the squared voltage function: $\int_{0}^{1} [V(t)]^2 dt = \int_{0}^{1} V_0^2 t^3 dt$ $= V_0^2 \int_{0}^{1} t^3 dt$ $= V_0^2 \left[ \frac{t^{3+1}}{3+1} \right]_{0}^{1}$ $= V_0^2 \left[ \frac{t^4}{4} \right]_{0}^{1}$ $= V_0^2 \left( \frac{1^4}{4} - \frac{0^4}{4} \right)$ $= V_0^2 \left( \frac{1}{4} - 0 \right) = \frac{V_0^2}{4}$

3. Calculate the RMS value: $V_{RMS} = \sqrt{\frac{1}{1} \times \frac{V_0^2}{4}}$ $V_{RMS} = \sqrt{\frac{V_0^2}{4}}$ $V_{RMS} = \frac{V_0}{2}$