Informative line

Average And Rms Value

Determine the root mean square value of alternating current, find the average and root mean square value of the given waveform in the time interval 0 to 2 sec. Learn alternating current definition with examples.

Alternating Current

When the direction of charges changes alternatively, then the current which flows in circuit is known as alternating current.

Symbol of Alternating Current

Alternating current is denoted by '$$\sim$$'

Sinusoidal Variation of Alternating Current with Time

An alternating current that varies sinusoidally with time is given by

$$i=i_0\,sin\,\omega t$$

where,

$$i_0$$ is the maximum value of current.

$$\omega$$ is the angular frequency of current.  Direct Current

When charges in a circuit flow in only one direction, then the current flowing is known as direct current.

Direction of Direct Current

Direct current flows in only one direction.

Symbol of Direct Current

The Direct current is denoted by  When the value of current does not vary with time, then such current is known as steady current.

If the current $$i$$ varies as a function $$i=i_0\,sin\,\omega t$$ , then which type of alternating current is this?

A B C D None of these

×

Option (A) is correct because current  $$i=i_0\,sin\,\omega t$$  is a function of sine, thus its graph will be: Option (B) & (C) are incorrect because given graphs do not describe the given equation of alternating current.

If the current $$i$$ varies as a function $$i=i_0\,sin\,\omega t$$ , then which type of alternating current is this?

A B C D

None of these

Option A is Correct

Calculation of Charge Flowing in a Circuit

• Let current is given as a function of time, $$i=f(t)$$  • The current is the flow of charges with time.

$$i.e.,$$  $$i=\dfrac{dq}{dt}$$

$$\Rightarrow dq=i\,dt$$

• The charge flowing in time interval from t1 to t2 is given by

$$q=\int\limits_{ t_1}^{t_2} i\,dt$$

If the current is given by $$I=I_0\,sin\,\omega t$$ , then calculate the charges flowing in half-cycle $$i.e.,$$ from 0 to $$\dfrac{T}{2}$$ time.

A Zero

B $$\dfrac{2I_0}{\omega}$$

C $$\dfrac{3I_0}{\omega}$$

D $$\dfrac{I_0}{\omega}$$

×

If the total time period is T then charges flowing in  $$0\,\,to\,\dfrac{T}{2}$$  time interval is given by

$$q=\int\limits_{ _0}^{T/_2} I\,dt$$

$$q=\int\limits_{ _0}^{T/_2} I_0\,sin \,\omega t\,dt$$

$$q=I_0\int\limits_{ _0}^{T/_2}sin \,\omega t\,dt$$

$$q=I_0 \left [ \dfrac{-cos\,\omega t}{\omega} \right]^{T/2}_0$$

$$q=\dfrac{-I_0}{\omega}\Big [cos\,\omega t\Big]_0^{T/2}$$

$$q=\dfrac{-I_0}{\omega}\left [cos\,\dfrac{ \omega \,T}{2}-cos \,0\right]$$

$$q=\dfrac{-I_0}{\omega}\left [cos\,\dfrac{2\pi}{T}\times\dfrac{T}{2}-1\right]$$

$$q=\dfrac{-I_0}{\omega}[-1\,-1]$$

$$q=\dfrac{-I_0}{\omega}[-2]$$

$$q=\dfrac{2I_0}{\omega}$$

If the current is given by $$I=I_0\,sin\,\omega t$$ , then calculate the charges flowing in half-cycle $$i.e.,$$ from 0 to $$\dfrac{T}{2}$$ time.

A

Zero

.

B

$$\dfrac{2I_0}{\omega}$$

C

$$\dfrac{3I_0}{\omega}$$

D

$$\dfrac{I_0}{\omega}$$

Option B is Correct

Average Value of i2 Over a Time Interval

The instantaneous current  $$\text i$$ can be positive or negative at a given instant, but the quantity $$i^2$$ will always remain positive, i.e. it is effective value of alternating current.

The average value of  $$\text i^2$$ over a time interval t1 to t2 is given by,

$$\overline {\text i^2}={{\dfrac{1}{t_2-t_1}}\,\int\limits^{t_2}_{t_1}\text i^2\,dt}$$

where,

$$\text i=f(t)$$

Root Mean Square Value of Alternating Current

The square root of mean square current is called root-mean-square current or rms current.

$$\text i_{rms}=\sqrt{ {\overline{\text i^2}}}$$

Root Mean Square Value

• Root mean square value is a mathematical quantity.
• It is used to compare alternating current and direct current.
• Root mean square value of the alternating current is the direct current which when passed through a resistor for a given period of time would produce the same heat as that produced by alternating current when passed through the same resistor for the same time.

Determine the root mean square value of alternating current $$I=I_0\,sin\,\omega t$$ for full cycle $$i.e.$$ from time interval $$0$$ to $$T$$.

A $$\dfrac{I_0}{2}$$

B $$\dfrac{I_0^2}{2}$$

C $$2\,I_0$$

D $$\dfrac{I_0}{\sqrt2}$$

×

Given : $$I= I_0\,sin\,\omega t$$

$$I^2= I_0{^2}\,sin^2\,\omega t$$

$$\overline{I^2}= \dfrac{1}{T}\,\int \limits ^T_0I_0{^2}\,sin^2\,\omega t\,dt$$

$$\overline{I^2}= \dfrac{I_0^2}{2\,T}\,\int \limits ^T_0\,(1-cos\,2\,\omega t)\,dt$$

$$\overline{I^2}= \dfrac{I_0^2}{2\,T}\,\left [t-\dfrac{sin\,2\,\omega t}{2\,\omega} \right]^T_0$$

$$\overline{I^2}= \dfrac{I_0^2}{2\,T}\,\left [T-\dfrac{sin\,2\,\omega \times T}{2\,\omega} -0\right ]$$

$$\overline{I^2}= \dfrac{I_0^2}{2\,T}\,\left [T-\dfrac{sin\,2\,\times{\dfrac{2\pi}{T}} \times T}{2\,\omega} -0\right ]$$

$$\overline{I^2}= \dfrac{I_0^2}{2\,T}[T]$$

$$\overline{I^2}= \dfrac{I_0^2}{2}$$

Root mean square value of given current

$$I_{rms}=\sqrt {\overline{I^2}}$$

$$I_{rms}=\sqrt {\dfrac{I^2_0}{2}}$$

$$I_{rms}={\dfrac{I_0}{\sqrt2}}$$

Determine the root mean square value of alternating current $$I=I_0\,sin\,\omega t$$ for full cycle $$i.e.$$ from time interval $$0$$ to $$T$$.

A

$$\dfrac{I_0}{2}$$

.

B

$$\dfrac{I_0^2}{2}$$

C

$$2\,I_0$$

D

$$\dfrac{I_0}{\sqrt2}$$

Option D is Correct

Root Mean Square Value of Alternating Current

• Consider a triangular wave of voltage which is given as follows :  By coordinate geometry, equation of straight line is given by

$$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$$ ....................(1)

where,

$$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ are end points of straight line.

For tine interval 0 to t1,

Putting the value of end points in equation .....(1)

let $$(x_1,\,y_1)=(0,\,0)\, \text{&}\,(x_2\,,y_2)=(t_1\,,V_o)$$

$$y-0=\dfrac{V_0-0}{t_1-0}(x-0)$$

$$y=\dfrac{V_0}{t_1}x$$

Here, $$y=V(t)\,\text&\,x=t$$

$$\Rightarrow V(t)=\left( \dfrac {V_0}{t_1} \right )t$$

Thus, the equation of voltage will be given as:

$$V(t)=\left( \dfrac {V_0}{t_1} \right )t$$  when, $$0\leq\,t\,\leq t_1$$

$$V(t)=0$$  when, $$t_1<\,t\,\leq T$$  • The average value of voltage is given by:

$$V_{avg}=\dfrac{1}{T}\displaystyle\int\limits ^T_0 V(t)\,dt$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left[\displaystyle\int\limits ^{t_1}_0 V_1(t)\,dt+\int\limits^T_{t_1}V_2(t)\,dt\right]$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left[\displaystyle\int\limits ^{t_1}_0 \dfrac{t}{t_1}V_0\,dt+0\right]$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left(\dfrac{V_0}{t_1} \right)\displaystyle\int\limits ^{t_1}_0 t\,dt$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left(\dfrac{V_0}{t_1} \right)\left[\dfrac{t^2}{2}\right]^{t_1}_0$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left(\dfrac{V_0}{t_1} \right)\left[\dfrac{t_1^2}{2}\right]$$

$$\Rightarrow V_{avg}=\dfrac{1}{T}\left[\dfrac{V_0\,t_1}{2} \right]$$

$$\Rightarrow V_{avg}=\dfrac{V_0\,t_1}{2\,T}$$

• Value of  $$\overline{V^2}$$ :

$$\overline{V_1^2}=\dfrac{1}{t_2-t_1}\,\displaystyle\int\limits ^{t_2} _{t_1}V_1^2(t)\,dt$$

$$\overline{V^2}=\dfrac{1}{t_1-0}\,\displaystyle\int\limits ^{t_1} _0\left(\dfrac{t^2}{t_1^2}\right)V_0^2\,dt$$

$$\overline{V_1^2}=\dfrac{V_0^2}{t_1(t_1^2)}\,\displaystyle\int\limits ^{t_1} _0t^2\,dt$$

$$\overline{V_1^2}=\dfrac{V_0^2}{t_1(t_1)^2}\,\left[\dfrac{t^3}{3}\right]^{t_1}_0$$

$$\overline{V_1^2}=\dfrac{V_0^2}{t_1(t_1)^2}\,\left[\dfrac{t_1^3}{3}\right]$$

$$\overline{V_1^2}=\dfrac{V_0^2}{3}$$

$$\overline{V_2^2}=\dfrac{1}{t_2-t_1}\,\displaystyle\int\limits ^{t_2} _{t_1}V_2^2(t)\,dt$$

$$\overline{V_2^2}=\dfrac{1}{T-t_1}\,\displaystyle\int\limits ^T _{t_1}0\,dt$$

$$\overline{V_2^2}=0$$

Thus,

$$\overline{V^2}=\overline{V_1^2}+\overline{V_2^2}$$

$$\overline{V^2}=\dfrac{V_0^2}{3}+0$$

$$\overline{V^2}=\dfrac{V_0^2}{3}$$

Root Mean Square Value

$$V_{rms}=\sqrt {\overline {V^2}}$$

$$V_{rms}={\sqrt{\dfrac {V_0^2}{3}}}$$

$$V_{rms}=\dfrac{V_0}{\sqrt3}$$

Find the average and root mean square value of the given waveform in the time interval 0 to 2 sec.

A $$V_{avg}=\dfrac{1}{2}volt,\,V_{rms}= \dfrac{2}{\sqrt3}\,volt$$

B $$V_{avg}=\dfrac{2}{3}volt,\,V_{rms}= 3\,volt$$

C $$V_{avg}=\dfrac{2}{\sqrt3}volt,\,V_{rms}= \dfrac{1}{2}\,volt$$

D $$V_{avg}=1volt,\,V_{rms}= 4\,volt$$

×

By coordinate geometry, equation of straight line is given by

$$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$$ ....................(1)

where,

$$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$ are end points of straight line. For the time interval 0 to 1 sec. -

Putting the value of end points in equation ...........(1)

let  $$(x_1,\,y_1)=(0,\,0)$$ & $$(x_2,\,y_2)=(1,\,2)$$

$$y-0=\dfrac{2-0}{1-0}(x-0)$$

$$y=\dfrac{2}{1}x$$

$$y=2x$$

Here, $$y=V(t) \,\text &\, x=t$$

$$\Rightarrow\,V(t)=2t$$  For the time interval 1 to 2 sec. -

From graph,

$$y=0$$

Therefore,  $$V(t)=0$$

for $$(1< t \leq 2)$$  The equation of voltage can be written as,

$$V_1(t)=2t$$  when, $$(0\leq t \leq 1)$$

$$V_2(t)=0$$   when, $$(1< t \le 2)$$ The average value of voltage is given by:

$$V_{avg}=\dfrac{1}{T}\displaystyle\int\limits ^T_0 V(t)\,dt$$

$$V_{avg}=\dfrac{1}{2}\displaystyle\int\limits ^2_0 V(t)\,dt$$

$$V_{avg}=\dfrac{1}{2}\left[\displaystyle\int\limits ^{1}_0 V_1(t)\,dt+\int\limits^2_{1}V_2(t)\,dt\right]$$

$$V_{avg}=\dfrac{1}{2}\left[\displaystyle\int\limits ^{1}_0 2\,t\,dt+0\right]$$

$$V_{avg}=\dfrac{1}{2} \left\{2\left[\dfrac{t^2}{2}\right]^1_0 \right\}$$

$$V_{avg}=\dfrac{1}{2} \left\{2\left[\dfrac{1}{2}\right]\right\}$$

$$V_{avg}=\dfrac{1}{2}\,Volt$$ • Value of  $$\overline{V^2}$$ :

$$\overline{V_1^2}=\dfrac{1}{t_2-t_1}\,\displaystyle\int\limits ^{t_2} _{t_1}V_1^2(t)\,dt$$

$$\overline{V_1^2}=\dfrac{1}{1-0}\,\displaystyle\int\limits ^{1} _0(2t)^2\,dt$$

$$\overline{V_1^2}=\displaystyle\int\limits ^{1} _04t^2\,dt$$

$$\overline{V_1^2}=4\left[\dfrac{t^3}{3}\right]^{1}_0$$

$$\overline{V_1^2}=\dfrac{4}{3}volt$$

$$\overline{V_2^2}=\dfrac{1}{t_2-t_1}\,\displaystyle\int\limits ^{t_2} _{t_1}V_2^2(t)\,dt$$

$$\overline{V_2^2}=\dfrac{1}{2-1}\,\displaystyle\int\limits ^2 _{1}0\,dt$$

$$\overline{V_2^2}=0$$

Thus,

$$\overline{V^2}=\overline{V_1^2}+\overline{V_2^2}$$

$$\overline{V^2}=\dfrac{4}{3}+0$$

$$\overline{V^2}=\dfrac{4}{3}volt$$ Root Mean Square Value

$$V_{rms}=\sqrt {\overline {V^2}}$$

$$V_{rms}=\sqrt{\dfrac{4}{3}}$$

$$V_{rms}=\dfrac{2}{\sqrt3}volt$$ Find the average and root mean square value of the given waveform in the time interval 0 to 2 sec. A

$$V_{avg}=\dfrac{1}{2}volt,\,V_{rms}= \dfrac{2}{\sqrt3}\,volt$$

.

B

$$V_{avg}=\dfrac{2}{3}volt,\,V_{rms}= 3\,volt$$

C

$$V_{avg}=\dfrac{2}{\sqrt3}volt,\,V_{rms}= \dfrac{1}{2}\,volt$$

D

$$V_{avg}=1volt,\,V_{rms}= 4\,volt$$

Option A is Correct

Average Value of Alternating Current

• Average current in any circuit is given by

$$I_{avg}=\dfrac{\Delta q}{\Delta t}$$

• $$\Delta q$$ is given by

$$\Delta q=\int\limits^{t_2}_{t_1} \,i\,dt$$  • In case of dc

$$I_{avg}=I_{dc}$$

$$\Delta q=I_{dc}\,\Delta t$$  Average Value of Alternating Current

• Average value of Alternating current is that dc value of current which will transfer same amount of charge as transferred by ac in same time interval.

since  $$\Delta t=t_2-t_1$$

$$\Rightarrow\,\Delta q=I_{dc\,}(t_2-t_1)$$

since  $$\Delta q=\int\limits^{t_2}_{t_1} \,i\,dt$$

$$\Rightarrow\,\int\limits^{t_2}_{t_1}i\,dt=I_{dc\,}(t_2-t_1)$$

$$\Rightarrow I_{dc}=\dfrac{1}{t_2-t_1\,}\int\limits^{t_2}_{t_1} \,i\,dt$$

$$I_{avg}=I_{dc}=\dfrac{1}{t_2-t_1\,}\int\limits^{t_2}_{t_1} \,i\,dt$$

If current flowing in a circuit is $$I=I_0\,sin\,\,\omega t$$, then calculate the average value or mean value of alternating current for full cycle i.e., from time interval $$0$$ to $$T$$.

A $$\dfrac{2\,I_0}{\omega}$$

B $$\dfrac{I_0}{\omega}$$

C Zero

D $$I_0$$

×

The graph of current $$I=I_0\,sin\,\omega t$$ is shown in figure

Here time period, $$T=\dfrac{2\pi}{\omega}$$ Average value of alternating current is given by

$$I_{avg}=\dfrac{1}{T}\int\limits^{T}_0 I\,dt$$

$$I_{avg}=\dfrac{1}{T}\int\limits^{T}_0 I_0\,sin\,\omega t\,dt$$

$$I_{avg}=\dfrac{I_0}{T}\int\limits^{T}_0 sin\,\omega t\,dt$$

$$I_{avg}=\dfrac{I_0}{T}\left [ \dfrac{-cos\,\omega t}{\omega} \right]^{T}_{0}$$

$$I_{avg}=\dfrac{-I_0}{T\omega}\left [{cos\,\omega t} \right]^{T}_{0}$$

$$I_{avg}=\dfrac{-I_0}{T\omega}\left [{cos\,\omega T-cos\,0^{\circ}}\right]$$

$$I_{avg}=\dfrac{-I_0}{T\omega}\left [cos\,\dfrac{2\pi}{T}\times T-1\right]$$

$$I_{avg}=\dfrac{-I_0}{T\omega}[1-1]$$

$$I_{avg}=0$$

If current flowing in a circuit is $$I=I_0\,sin\,\,\omega t$$, then calculate the average value or mean value of alternating current for full cycle i.e., from time interval $$0$$ to $$T$$.

A

$$\dfrac{2\,I_0}{\omega}$$

.

B

$$\dfrac{I_0}{\omega}$$

C

Zero

D

$$I_0$$

Option C is Correct

Concept of Heat

• Consider a resistor of resistance R which is carrying a current,

$$i=i_0\,sin\,\omega t$$

Heat Developed Over a Time Period

Heat developed over a time period is given by

$$U=\int \limits^T_0i^2R\,dt$$

$$U=\int \limits^T_0(i_0^2\,sin^2\,\omega t)R\,dt$$

$$U=R\,i_0^2\,\int \limits^T_0\,sin^2\,\omega t\,dt$$

$$U=\dfrac{i_0^2\,R}{2}\,\int \limits^T_0\,(1-cos\,2\,\omega t)\,dt$$

$$U=\dfrac{i_0^2R}{2}\,\left[t-\dfrac{sin\,2\,\omega t}{2\,\omega}\,\right]^T_0$$

$$U=\dfrac{i_0^2R}{2}\,\left[T-\dfrac{sin\,2 \times\dfrac{2\pi}{T}\times T}{2\omega}-0\,\right]$$

$$Heat=U=\dfrac{i_0^2\,R\,T}{2}$$

If given current is $$I=5\,sin\,\omega t$$ and $$R=2\,\Omega$$ then find the heat generated iin $$0\,\, to\,\, 2\, sec$$ time interval.

A 50 J

B 70 J

C 60 J

D 80 J

×

Amplitude, $$I_0=5$$

Time, $$T=2\,sec$$

Resistance,  $$R=2\,\Omega$$

Heat generated is

$$U=\dfrac{I_0^2\,R\,T}{2}$$

$$U=\dfrac{25\times2\times2}{2}$$

$$U=50\,J$$

If given current is $$I=5\,sin\,\omega t$$ and $$R=2\,\Omega$$ then find the heat generated iin $$0\,\, to\,\, 2\, sec$$ time interval.

A

50 J

.

B

70 J

C

60 J

D

80 J

Option A is Correct