Root Mean Square Value of A.C. and A.C. Through a Resistance Only

Root Mean Square (R.M.S) or Virtual Value of A.C. or Effective Value of A.C.

It is that steady current when passed through a resistance for a given time produces the same amount of heat as the alternating current does in the same resistance in the same time.

To calculate its value, let an alternating current be represented by

$$ I = I_0 \sin \omega t $$

If this alternating current flows through a resistance R for small time dt, then small amount of heat produced which is given by

\begin{align*} dH &= I^2R\: dt = (I_0^2\sin^2\omega t) R\: dt \\ \text {or,} \: dH &= I_o^2 R \sin^2\omega T dt \\ \end{align*}

To obtain total amount og heat produced in one cycle, we have to integrate equation (i) from t = 0 to t = T.

\begin{align*}&= I_o^2 R \int_0^T \sin ^2 \omega t \: dt \\ &= I_o^2 R \int _0^T \frac {(1 - \cos 2\omega t)}{2} dt \\ &= \frac {I_0^2R}{2}\int _0^T (1 - \cos 2 \omega t) dt \\ &= \frac {I_0^2R}{2} \left [ \int _0^T dt - \int _0^T \cos 2\omega t dt \right ] \\&= \frac {I_0^2R}{2} \left [ [\text {t}]_0^T - \left [\frac {\sin 2\omega t}{2\omega } \right ]_0^T\right ] \\ &= \frac {I_0^2R}{2}\left [(T-0)-\frac {1}{2\omega}\left [\sin 2\frac {2\pi}{T.}t \right ]_0^T \right ] \\ &= \frac {I_0^2R}{2} \left [ T - \frac {1}{2\omega } \left (\sin 2 \frac {2\pi }{T}T - \sin 0 \right )\right ] \\ \text {or,} \: H &= \frac {I_0^2 RT}{2} \dots (ii) \\ \end{align*}

If I_v be the r.m.s. value or virtual value of a.c., then the amount of heat produced in the same resistance R in same time T is written

\begin{align*} H &= I_v^2 RT \dots (ii) \\ \text {From equation} \: (ii) \: \text {and} \: (iii), \text {we get} \\ I_v^2RT &= \frac {I_0^2 RT}{2} \\ \therefore \: I_v &= \frac {I_0}{\sqrt {2}} = 0.707\: I_0 \\ \end{align*}

Hence the r.m.s. value or virtual value or effective value of a.c. in 0.707 times the peak value of a.c. i.e 70.7% of the peak value of a.c.

The virtual value of alternating current can be proved as

$$ E_v = \frac {E_0}{\sqrt 2} = 0.707\: E_0 $$

A.C. Through a Resistance Only

An a.c. source is connected to a resistor is known as resistive circuit.

\begin{align*} \text {The applied alternating e.m.f. is given by} \\ E &= E_0 \sin \omega t \\ \text {Let I be the circuit at any instant of time t. So} \\\text {The potential difference across the resistor} &= IR \\ \text {or,} \: E &= IR \\ \text {or,} \: I &= \frac ER \\ &= \frac {E_0 \sin \omega t }{R}\\ [\therefore E = E_0 \sin \omega t \: \text {in equation} \: (i)] \\\text {or,} \: I &= I_0 \sin \omega T \dots (ii)\\ \text {where} \: \frac {E_0}{R} = I_0 \: \text {is the peak value of alternating current.} \end{align*}

Comparing equation (i) and (ii), we find that E and I are in phase. Therefore, in an a.c. circuit containing R only, the current and e.m.f. are in the phase.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel.Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.