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