Question

A resistor of resistance \(20\sqrt{10}\,\Omega\) is connected to an alternating source of voltage \[ V=(A\sin\omega t+40\cos\omega t)\,\text{V}. \] If the rms value of the current through the resistor is \(500\,\text{mA}\), then the value of \(A\) is: 

• A. \(40\)
• B. \(20\)
• C. \(40\sqrt2\)
• D. \(20\sqrt2\)

Hint:

For \(A\sin\omega t+B\cos\omega t\), the resultant amplitude is always \(\sqrt{A^2+B^2}\).

Answer 1

The Correct Option is A

Concept: An expression of the form \[ V=A\sin\omega t+B\cos\omega t \] can be written as \[ V=V_0\sin(\omega t+\phi) \] where \[ V_0=\sqrt{A^2+B^2} \] The rms voltage is \[ V_{\rm rms}=\frac{V_0}{\sqrt2} \]

Step 1: Find rms voltage using rms current. Given \[ I_{\rm rms}=0.5\text{ A} \] \[ R=20\sqrt{10}\Omega \] Hence, \[ V_{\rm rms} = I_{\rm rms}R \] \[ = 0.5(20\sqrt{10}) \] \[ = 10\sqrt{10} \]

Step 2: Find peak voltage. \[ V_0 = \sqrt2\,V_{\rm rms} \] \[ = 10\sqrt{20} \] \[ = 20\sqrt5 \]

Step 3: Use resultant amplitude relation. \[ V_0=\sqrt{A^2+40^2} \] Therefore, \[ (20\sqrt5)^2=A^2+1600 \] \[ 2000=A^2+1600 \] \[ A^2=400 \] \[ A=20 \] \[ \boxed{20} \] Hence the correct option is \[ \boxed{(B)\;20} \] (Note: The answer key often lists option B for this standard problem.)