Question

An ac voltage is given as \(v = 14\sin(314t)\,\text{V\). The average and the effective value of the voltage (in V) over a cycle are respectively :}

• A. \(14\) and \(7\)
• B. \(10\) and \(14\)
• C. \(0\) and \(10\)
• D. \(10\) and \(0\)

Hint:

For any sinusoidal alternating voltage or current: \[ V_{\text{avg}}=0 \] (over a complete cycle) and \[ V_{\text{rms}}=\frac{V_0}{\sqrt2}. \] Always remember the factor \(\sqrt2\) while converting peak values into rms values.

Answer 1

The Correct Option is C

Concept: An alternating voltage varies sinusoidally with time and is generally represented as \[ v = V_0 \sin \omega t \] where \[ V_0 = \text{peak voltage (amplitude)} \] and \[ \omega = \text{angular frequency}. \] For a sinusoidal alternating voltage: \[ V_{\text{avg}} = 0 \] over one complete cycle because the positive half-cycle and negative half-cycle are equal in magnitude and opposite in sign. The rms (root mean square) value or effective value is given by \[ V_{\text{rms}}=\frac{V_0}{\sqrt{2}}. \] The rms value represents the dc voltage that would produce the same heating effect in a resistor.

Step 1: Identify the peak voltage. The given alternating voltage is \[ v=14\sin(314t). \] Comparing with \[ v=V_0\sin\omega t, \] we obtain \[ V_0=14\,\text{V}. \] Thus, the peak voltage is \[ 14\,\text{V}. \]

Step 2: Find the average value over one complete cycle. For a complete cycle of a sine wave, \[ V_{\text{avg}}=0. \] This is because the positive and negative halves cancel each other exactly. Hence, \[ V_{\text{avg}}=0\,\text{V}. \]

Step 3: Calculate the rms (effective) value. Using \[ V_{\text{rms}} = \frac{V_0}{\sqrt2}, \] we get \[ V_{\text{rms}} = \frac{14}{\sqrt2}. \] Multiplying numerator and denominator by \(\sqrt2\), \[ V_{\text{rms}} = \frac{14\sqrt2}{2} = 7\sqrt2. \] Using \[ \sqrt2 \approx 1.414, \] \[ V_{\text{rms}} = 7\times1.414 = 9.898. \] Therefore, \[ V_{\text{rms}}\approx10\,\text{V}. \]

Step 4: Write the final answer. Thus, \[ V_{\text{avg}}=0\,\text{V} \] and \[ V_{\text{rms}}=10\,\text{V}. \] Therefore, \[ \boxed{\text{(C) }0\text{ and }10} \] is the correct answer.