
In a circuit, the current lags behind the voltage by a phase difference of \[{\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}\], the circuit will contain which of the following:
A. Only R
B. Only C
C. R and C
D. Only L
Answer
565.5k+ views
Hint: We will analyse the pure inductor, purely capacitor and purely resistor circuit to find the relation between current and voltage. We will use the concept of Ohm’s law which say that voltage through a circuit is equal to the product of current and resistance of the circuit.
Complete step by step answer:Let us consider a resistor of resistance R is connected with to an a.c supply whose voltage can be written as below:
\[E = {E_0}\sin \omega t\]……(1)
Here \[{E_0}\] is the maximum value of voltage.
We know that the voltage of a resistor circuit is equal to the product of current and resistance.
\[E = IR\]
And,
\[{E_0} = {I_0}R\]
Here I is the value of current in the circuit and \[{I_0}\] is the maximum current in the circuit.
We will substitute \[{I_0}R\] for \[{E_0}\] and \[IR\] for E in equation (1).
\[\begin{array}{c}
IR = {I_0}R\sin \omega t\\
I = {I_0}\sin \omega t
\end{array}\]……(2)
From equation (1) and equation (2), we can conclude that current is in phase with voltage in case of a purely resistive circuit.
Let us consider that the expression for voltage in case of the purely capacitive and inductive circuit is the same as that in case of a purely resistive circuit.
We know that the below expression gives the relation for current in purely capacitive circuit:
\[I = {I_0}\sin \left( {\omega t + {\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}} \right)\]……(3)
From equation (1) and equation (3), we can conclude that the current in a purely capacitive circuit leads by a phase angle of \[{\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}\].
We know that in the case of an inductor, the expression of current can be written as:
\[I = {I_0}\sin \left( {\omega t - {\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}} \right)\]……(4)
Therefore, based on equation (1) and equation (4), we can conclude that the current lags by a phase difference \[{\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}\] in the case of a purely inductive circuit.
Hence, the option (D) is correct.
Note:The expression for the current in purely capacitive circuits comes in the form of cosine. To compare it with the expression of voltage to find the phase difference, we converted it into the form of sine.
Complete step by step answer:Let us consider a resistor of resistance R is connected with to an a.c supply whose voltage can be written as below:
\[E = {E_0}\sin \omega t\]……(1)
Here \[{E_0}\] is the maximum value of voltage.
We know that the voltage of a resistor circuit is equal to the product of current and resistance.
\[E = IR\]
And,
\[{E_0} = {I_0}R\]
Here I is the value of current in the circuit and \[{I_0}\] is the maximum current in the circuit.
We will substitute \[{I_0}R\] for \[{E_0}\] and \[IR\] for E in equation (1).
\[\begin{array}{c}
IR = {I_0}R\sin \omega t\\
I = {I_0}\sin \omega t
\end{array}\]……(2)
From equation (1) and equation (2), we can conclude that current is in phase with voltage in case of a purely resistive circuit.
Let us consider that the expression for voltage in case of the purely capacitive and inductive circuit is the same as that in case of a purely resistive circuit.
We know that the below expression gives the relation for current in purely capacitive circuit:
\[I = {I_0}\sin \left( {\omega t + {\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}} \right)\]……(3)
From equation (1) and equation (3), we can conclude that the current in a purely capacitive circuit leads by a phase angle of \[{\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}\].
We know that in the case of an inductor, the expression of current can be written as:
\[I = {I_0}\sin \left( {\omega t - {\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}} \right)\]……(4)
Therefore, based on equation (1) and equation (4), we can conclude that the current lags by a phase difference \[{\pi {\left/
{\vphantom {\pi 2}} \right.
} 2}\] in the case of a purely inductive circuit.
Hence, the option (D) is correct.
Note:The expression for the current in purely capacitive circuits comes in the form of cosine. To compare it with the expression of voltage to find the phase difference, we converted it into the form of sine.
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