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.
By Kirchhoff's law :-
\(V_0 \,sin\,\omega t - IR = 0\)
\(V_0 \,sin\,\omega t = IR \\ I=\dfrac{V_0}{R}\,sin\,\omega t\)
\(V = V_0 \,sin\, \omega t\)
Note : Both current and voltage are functions of sine. Thus, both will increase or decrease simultaneously.
A \(8\, sin\, \omega t\)
B \(16\, sin\, \omega t\)
C \(2\, sin\, \omega t\)
D \(10\, sin\, \omega t\)
By Kirchoff's law-
\(V_0 \,sin \,\omega t - iR = 0\)
\(V_0 \,sin\,\omega t = iR \)
\(i=\dfrac{V_0}{R}sin\,\omega t\)
\(V = V_0 \,sin\,\omega t\)
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
Consider two bulbs B_{1} and B_{2 }of different ratings (P_{1}, V_{1}) and (P_{2}, V_{2}) respectively, are connected in series and operated at frequency \(\text f\).
The AC voltage is \(V=V_0\,sin \,\omega t\).
where,
P_{1 }\({\to}\) Power of Bulb B_{1}
V_{1 }\({\to}\) Voltage of Bulb B_{1}
f \({\to}\) Frequency of Bulb B_{1} and B_{2}
P_{2 }\({\to}\) Power of Bulb B_{2}
V_{2 }\({\to}\) Voltage of Bulb B_{2}
For bulb B_{1 }: Calculate, R1 = \(\dfrac{V_1^2}{P_1}\)
For bulb B_{2}_{ }: Calculate, R_{2} = \(\dfrac{V_2^2}{P_2}\)
R_{eq }= R_{1} + R_{2 }
and calculate : \(\text I_0=\dfrac{V}{R_{eq}}\)
\(\text I_{rms}\)= \(\dfrac{I_0}{\sqrt2}\)
A \(\sqrt 2\, mA\)
B \(3\sqrt2\, mA\)
C \(10\,mA\)
D \(\dfrac {3}{2}\,mA\)
where,
P_{1 }\({\to}\) Power of Bulb B_{1}
V_{1 }\({\to}\) Voltage of Bulb B_{1}
f \({\to}\) Frequency of Bulb B_{1} and B_{2}
P_{2 }\({\to}\) Power of Bulb B_{2}
V_{2 }\({\to}\) Voltage of Bulb B_{2}
To calculate \(\text I_{rms}\)_{, }following steps are to be followed :
For bulb B_{1 }: calculate, R_{1} = \(\dfrac{V_1^2}{P_1}\)
For bulb B_{2}_{ }: calculate, R_{2} = \(\dfrac{V_2^2}{P_2}\)
\(\dfrac{1}{R_{eq}}\) = \(\dfrac{1}{R_1}\) + \(\dfrac{1}{R_2}\)
and calculate :_{ \(\text I_0=\dfrac{V_0}{R_{eq}}\)}
\(\text I_{rms}\) = \(\dfrac{I_0}{\sqrt2}\)
A \(\dfrac{1}{\sqrt2}\) A
B \(\dfrac{1}{2\,\sqrt2}\) A
C \(\dfrac{1}{3\,\sqrt2}\) A
D \(\dfrac{1}{4\,\sqrt2}\) A
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.
The current-time and voltage-time graph are as follows :
(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.
I_{rms} = \(\dfrac{I_0}{\sqrt2}\)
where, \(I_0\) is maximum value of current in the circuit.
where, V_{0 }is maximum value of voltage in the circuit.
\(\text I_{rms}\)_{ }= \(\dfrac{V_{rms}}{R}\)