Question

Voltage (V) equation in an alternating current circuit is represented by \( V = 40 \sin(100 \pi t) \) volt. Here t is in second. Draw the time-voltage graph for one cycle with proper scale.

Hint:

When asked to draw a sinusoidal graph, always calculate the amplitude and the time period first. Then, mark the key points at t=0, T/4, T/2, 3T/4, and T to ensure the shape and scale of the sine wave are correct.

Answer 1

Step 1: Understanding the Concept: 
The given equation is for a sinusoidal alternating voltage. We need to identify its key characteristics (amplitude, frequency, time period) to plot its graph accurately for one complete cycle. 
Step 2: Key Formula or Approach: 
The standard equation for a sinusoidal AC voltage is \( V(t) = V_m \sin(\omega t) \), where:
\( V_m \) is the peak voltage or amplitude.
\( \omega \) is the angular frequency.
The time period (T) is related to \( \omega \) by \( T = \frac{2\pi}{\omega} \). 
Step 3: Detailed Explanation: 
Comparing the given equation \( V = 40 \sin(100 \pi t) \) with the standard form:
Peak Voltage (\(V_m\)): \( V_m = 40 \) V. This means the voltage varies between +40 V and -40 V.
Angular Frequency (\(\omega\)): \( \omega = 100\pi \) rad/s.
Now, we calculate the time period (T) for one complete cycle: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100\pi} = \frac{2}{100} = 0.02 \, \text{seconds} \] To draw the graph, we need some key points in the first cycle (from t=0 to t=0.02 s):
At \( t=0 \), \( V = 40\sin(0) = 0 \) V.
At \( t = T/4 = 0.005 \) s, \( V = 40\sin(100\pi . 0.005) = 40\sin(\pi/2) = 40 \) V (Peak).
At \( t = T/2 = 0.01 \) s, \( V = 40\sin(100\pi . 0.01) = 40\sin(\pi) = 0 \) V.
At \( t = 3T/4 = 0.015 \) s, \( V = 40\sin(100\pi . 0.015) = 40\sin(3\pi/2) = -40 \) V (Trough).
At \( t = T = 0.02 \) s, \( V = 40\sin(100\pi . 0.02) = 40\sin(2\pi) = 0 \) V.
Step 4: Graph: