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Ac With Resistance

Learn to analysis of circuit containing a resistor with AC source, and phasor diagram of voltage & current. Find the root mean square value of current when resistor of 100 ? resistance is connected across the source.

Analysis of Circuit Containing a Resistor with AC Source

  • Consider a circuit in which an AC voltage source is connected.
  • The voltage of AC voltage source is  \(V=V_0\, sin\, \omega t\) .
  • It is connected across a resistor.
  • Such a circuit is also known as a purely resistive circuit.

Current across the Resistor

By Kirchhoff's law :-

\(V_0 \,sin\,\omega t - IR = 0\)

\(V_0 \,sin\,\omega t = IR \\ I=\dfrac{V_0}{R}\,sin\,\omega t\)

Voltage Source

\(V = V_0 \,sin\, \omega t\)

Note : Both current and voltage are functions of sine. Thus, both will increase or decrease simultaneously.

Illustration Questions

If voltage source of voltage \( V = 4\, sin\, \omega t\)  is connected across a resistor of resistance \(2\,\Omega\), then find the current in the circuit as a function of time.

A \(8\, sin\, \omega t\)

B \(16\, sin\, \omega t\)

C \(2\, sin\, \omega t\)

D \(10\, sin\, \omega t\)

×

Voltage, \(V=4\, sin\, \omega t\)

Resistance, \(R=2\,\Omega\)

\(\text I\) = \(\dfrac{V}{R}\)

\(\text I\) = \(\dfrac{V_0\,sin\,\omega t}{R}\)

\(\text I\) = \(\dfrac{4\,sin\,\omega t}{2}\)

\(\text I\) =  \(2\, sin\, \omega t\)

If voltage source of voltage \( V = 4\, sin\, \omega t\)  is connected across a resistor of resistance \(2\,\Omega\), then find the current in the circuit as a function of time.

A

\(8\, sin\, \omega t\)

.

B

\(16\, sin\, \omega t\)

C

\(2\, sin\, \omega t\)

D

\(10\, sin\, \omega t\)

Option C is Correct

Graph of Voltage and Current against Time for Purely Resistive Circuit

  • Consider a circuit in which an AC voltage source is connected.
  • The voltage of AC voltage source is \(V=V_0\,sin\,\omega t\).
  • It is connected across a resistor.
  • Such a circuit is also known as a purely resistive circuit.

Current across the Resistor

By Kirchoff's law-

\(V_0 \,sin \,\omega t - iR = 0\)

\(V_0 \,sin\,\omega t = iR \)

\(i=\dfrac{V_0}{R}sin\,\omega t\)

Voltage Source

\(V = V_0 \,sin\,\omega t\)

Note : Both current and voltage are functions of sine. Thus, both will increase or decrease simultaneously.

Graph of Voltage against Time

Graph of Current against Time

Illustration Questions

If graph of voltage against time is as shown in figure then, which one of the following will be the graph of current against time when R = \(5\,\Omega\)?

A

B

C

D

×

Option (A) is correct because the phase of current is same as the phase of voltage and amplitude of current waveform is 1 A.

\(\text I\) = \(\dfrac{V_0\,sin\,\omega t}{R}\) 

\(\text I\) = \(\dfrac{5\,sin\,\omega t}{5}\)

\(\therefore\)\(\text I\) = \(1\,sin\,\omega t\)

Option(B), (C) and (D) are incorrect because the phase of current is not similar to the phase of given voltage.

If graph of voltage against time is as shown in figure then, which one of the following will be the graph of current against time when R = \(5\,\Omega\)?

image
A image
B image
C image
D image

Option A is Correct

Phasor Diagram of Voltage and Current

Phasor 

  • Phasor is a vector which rotates counterclockwise about the origin with angular frequency \(\omega\).

Length of Phasor

  • The length of the phasor is the maximum value of the quantity.
  • Vis the length of phasor.
  • \(V_0 \,sin \,\omega t\)  is the projection on vertical axis. It is an instantaneous value.

Use of Phasor 

  • Phasor are useful to understand the phase difference between different quantities.
  • If \(V\) is leading \(\text I\) by 900.

Analysis of Circuit containing a Resistor with AC Source

  • Consider a circuit in which an AC voltage source is connected.
  • The voltage of AC voltage source is  \(V=V_0\, sin\, \omega t\).
  • It is connected across a resistor.
  • Such a circuit is also known as a purely resistive circuit.

Current across the Resistor

By Kirchhoff's law :-

\(V_0 \,sin\,\omega t - IR = 0\)

\(V_0\, sin\,\omega t = IR \)

\(I=\dfrac{V_0}{R}\,sin\,\omega t\)

Voltage Source

\(V = V_0\, sin\,\omega t\)

Note : Both current and voltage are functions of sine. Thus, both will increase or decrease simultaneously.

Analyzing the graph of current and voltage against time

The current-time and voltage-time graph are as follows :

  • By considering the current-time and voltage-time graphs on same axis, following observations can be made :

(i) Along the curve O to P, the voltage increases and current also increases.

(ii) Along the curve P to Q, the voltage decreases and current also decreases.

(iii) Along the curve Q to R, the voltage decreases and current also decreases.

(iv) Along the curve R to S, the voltage increases and current also increases.

  • From above observations, it can be concluded that the current and voltage vary identically with time and attain same value at same time.
  • It means that current and voltage are in same phase.
  • The phasor diagram for the resistive circuit is shown in figure:

Illustration Questions

The voltage phasor for voltage \(V=4\,sin\,\omega t\) and resistor of resistance \(R=2\,\Omega\) is shown in figure. Choose the correct phasor of current.

A

B

C

D

×

Option A is correct because voltage and current are in same phase and amplitude of current is 2 A.

\(\text I_0\) = \(\dfrac{V_0}{R}\)

\(\text I_0\) = \(\dfrac{4}{2}=2\,A\)

 

Option (B), (C) and (D) are incorrect because the phase of current is not similar to the phase of given voltage.

The voltage phasor for voltage \(V=4\,sin\,\omega t\) and resistor of resistance \(R=2\,\Omega\) is shown in figure. Choose the correct phasor of current.

image
A image
B image
C image
D image

Option A is Correct

Root Mean Square Value of Current

Irms\(\dfrac{I_0}{\sqrt2}\)

where, \(I_0\) is maximum value of current in the circuit.

Root Mean Square Value of Voltage 

  • Vrms \(\dfrac{V_0}{\sqrt2}\)

where, Vis maximum value of voltage in the circuit.

  • Relation between Vrms and \(\text I_{rms}\) :

\(\text I_{rms}\) \(\dfrac{V_{rms}}{R}\)

Illustration Questions

The root mean square value of voltage is 141 V for an AC source of voltage \(V=V_0\,sin\,\omega t\) . Find the root mean square value of current when resistor of 100 \(\Omega\) resistance is connected across the source.

A 1.41 A

B 2 A

C 3 A

D 144 A

×

Vrms of AC source = 141 V

Resistance of resistor, R = 100 \(\Omega\)

 

\(\text I_{rms}\) =  \(\dfrac{V_{rms}}{R}\)

\(\text I_{rms}\) = \(\dfrac{141}{100}\)

\(\text I_{rms}\) = \(1.41\,A\)

 

The root mean square value of voltage is 141 V for an AC source of voltage \(V=V_0\,sin\,\omega t\) . Find the root mean square value of current when resistor of 100 \(\Omega\) resistance is connected across the source.

A

1.41 A

.

B

2 A

C

3 A

D

144 A

Option A is Correct

Root Mean Square Value of Current when Two Bulbs are Connected in Series

Consider two bulbs B1 and Bof different ratings (P1, V1) and (P2, V2) respectively, are connected in series and operated at frequency \(\text f\).

The AC voltage is \(V=V_0\,sin \,\omega t\).

where,

P\({\to}\) Power of Bulb B1

V\({\to}\) Voltage of Bulb B1

\({\to}\) Frequency of Bulb B1 and B2

P\({\to}\) Power of Bulb B2

V\({\to}\) Voltage of Bulb B2

Steps to Calculate  \(\text I_{rms}\) 

  • To calculate Irms , following steps are to be followed :

For bulb B1 : Calculate,  R1 = \(\dfrac{V_1^2}{P_1}\)

For bulb B2 : Calculate,  R2 = \(\dfrac{V_2^2}{P_2}\)

  • Calculate equivalent Resistance :

Req = R1 + R

  • Use formula :   \(\text I\) = \(\dfrac{V_0\,sin\,\omega t}{R_{eq}}\)

and calculate : \(\text I_0=\dfrac{V}{R_{eq}}\)

  • Calculate \(\text I_{rms}\) 

\(\text I_{rms}\)\(\dfrac{I_0}{\sqrt2}\) 

 

 

Illustration Questions

Two bulbs B1 and B2  of different ratings (60 W, 120 V) and (40 W, 120 V)  respectively, are connected in series and operated at 60 Hz. If the AC voltage is V =  \(6\sqrt2 \,sin\,\omega t\), then calculate \(\text I_{rms}\) for the combination. 

A \(\sqrt 2\, mA\)

B \(3\sqrt2\, mA\)

C \(10\,mA\)

D \(\dfrac {3}{2}\,mA\)

×

Power of Bulb B1, P1 = 60 W

Voltage of Bulb B1, V1 = 120 V

Frequency of Bulb B1, f1 = 60 Hz

Power of Bulb B2, P2 = 40 W

Voltage of Bulb B2, V2 = 120 V

Frequency of Bulb B2, f2 = 60 Hz

Voltage, V =  \(6\sqrt2 \,sin\,\omega t\)

For Bulb B1 : Resistance, R\(\dfrac{V_1^2}{P_1}\) = \(\dfrac{(120)^2}{60}\) = 240 \(\Omega\)

For Bulb B2 : Resistance, R=  \(\dfrac{V_2^2}{P_2}\) = \(\dfrac{(120)^2}{40}\) = 360 \(\Omega\)

 

Equivalent resistance, R = R1 + R2

= 240 + 360

= 600 \(\Omega\)                                                                                 

image

AC voltage, V =  \(6\sqrt2 \,sin\,\omega t\)

Resistance, R = 600 \(\Omega\)

Thus,

\(\text I_0\) \(\dfrac{V_0}{R}\)

 \(\text I_0\) = \(\dfrac{6\,\sqrt2}{600}\) 

\(\dfrac{\sqrt2}{100}\)

Root mean square value of current :

\(\text I_{rms}\) \(\dfrac{I_0}{\sqrt2}\) 

\(\dfrac{1\,\sqrt2}{100\,\sqrt2}\)

= \(0.01 A\)

= \(10\, mA\)

Two bulbs B1 and B2  of different ratings (60 W, 120 V) and (40 W, 120 V)  respectively, are connected in series and operated at 60 Hz. If the AC voltage is V =  \(6\sqrt2 \,sin\,\omega t\), then calculate \(\text I_{rms}\) for the combination. 

image
A

\(\sqrt 2\, mA\)

.

B

\(3\sqrt2\, mA\)

C

\(10\,mA\)

D

\(\dfrac {3}{2}\,mA\)

Option C is Correct

Root Mean Square Value of Current when Two Bulbs are Connected in Parallel

  • Consider two bulbs B1 and B2 of different ratings (P, V1) and (P, V2) respectively are connected in parallel and operated at f = 60 Hz. The AC voltage is  \(V=V_0\,sin\,\omega t\) .

where,

P\({\to}\) Power of Bulb B1

V\({\to}\) Voltage of Bulb B1

\({\to}\) Frequency of Bulb B1 and B2

P\({\to}\) Power of Bulb B2

V\({\to}\) Voltage of Bulb B2

Steps to calculate \(\text I_{rms}\) 

  • To calculate \(\text I_{rms}\)following steps are to be followed :

For bulb B: calculate, R1 = \(\dfrac{V_1^2}{P_1}\)

For bulb B2 : calculate, R2 = \(\dfrac{V_2^2}{P_2}\)

  • Calculate equivalent resistance :

\(\dfrac{1}{R_{eq}}\) = \(\dfrac{1}{R_1}\) + \(\dfrac{1}{R_2}\)

  • Use formula : \( \text I=\dfrac{V_0\,sin\,\omega t}{R_{eq}}\)

     and calculate : \(\text I_0=\dfrac{V_0}{R_{eq}}\)

  • Calculate \(\text I_{rms}\) :

\(\text I_{rms}\) = \(\dfrac{I_0}{\sqrt2}\)

 

Illustration Questions

Two bulbs B1 and B2 of different ratings (60 W, 120 V) and (60 W, 120 V) respectively are connected in parallel and operated at 60 Hz. If the AC voltage is V = 120 \(sin\,\omega t\) , then calculate \(\text I_{rms}\) for the combination.

A \(\dfrac{1}{\sqrt2}\) A

B \(\dfrac{1}{2\,\sqrt2}\) A

C \(\dfrac{1}{3\,\sqrt2}\) A

D \(\dfrac{1}{4\,\sqrt2}\) A

×

Power of Bulb B1, P1 = 60 W

Voltage of Bulb B1, V1 = 120 V

Power of Bulb B2, P2 = 60 W

Voltage of Bulb B2, V2 = 120 V

AC voltage, V = 120 \(sin\,\omega t\)

For Bulb B: Resistance, R\(\dfrac{V_1^2}{P_1}\) = \(\dfrac{(120)^2}{60}\) = 240 \(\Omega\)

For Bulb B2 : Resistance, R=  \(\dfrac{V_2^2}{P_2}\) = \(\dfrac{(120)^2}{60}\) = 240 \(\Omega\)

Equivalent Resistance :

\(\dfrac{1}{R_{eq}}\) = \(\dfrac{1}{R_1}\) + \(\dfrac{1}{R_2}\)

\(\dfrac{1}{R_{eq}}\) = \(\dfrac{1}{240}\) + \(\dfrac{1}{240}\)

\(\dfrac{1}{R_{eq}}\) = \(\dfrac{2}{240}\)

\({R_{eq}}=120 \,\Omega\)

image

\(I_0\) = \(\dfrac{V_0}{R}\)

\(I_0\) = \(\dfrac{120}{120}\) 

    \(= 1 \,A\)

Root mean square value of current 

\(I_{rms}\) = \(\dfrac{I_0}{\sqrt2}\) 

= \(\dfrac{1}{\sqrt2}\) \(A\)

Two bulbs B1 and B2 of different ratings (60 W, 120 V) and (60 W, 120 V) respectively are connected in parallel and operated at 60 Hz. If the AC voltage is V = 120 \(sin\,\omega t\) , then calculate \(\text I_{rms}\) for the combination.

image
A

\(\dfrac{1}{\sqrt2}\) A

.

B

\(\dfrac{1}{2\,\sqrt2}\) A

C

\(\dfrac{1}{3\,\sqrt2}\) A

D

\(\dfrac{1}{4\,\sqrt2}\) A

Option A is Correct

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