The phase relationship between current and voltage in a pure resistive circuit is best represented by.
A.
B.
C.
D .

Hint: A pure resistive circuit consists of an AC source and a resistor. There is no phase difference between voltage and current and the current and voltage is said to be in-phase.


Complete step by step answer:
It is given that the circuit is pure resistive which means that there is a resistor in the circuit along with an AC source.

Here ${\text{I}}$ is the current through the circuit and AC source has ${{\text{V}}_{\text{0}}}$ is the voltage difference and ${\text{R}}$ is resistance of the circuit.
Apply Kirchhoff’s law in the given circuit.

Appling Kirchhoff’s law,
$
  {V_0}\sin \omega t - I \cdot R = 0 \\
  I = \left( {\dfrac{{{V_0}}}{R}} \right)\sin \omega t \\
 $
So, here we can see that current in the same phase as of voltage but is less in magnitude then voltage.
The phase diagram in which both voltage and current are in the same phase and also the current is having lesser value than voltage. Let us observe each of the options and see which one is the correct answer for this problem.
In option A the value of current is lesser than that of voltage but there is phase difference in the diagram.
In option B the value of current is lesser than that of voltage but there is a phase difference that is present in the phase diagram.
In option C the value of current is lesser than that of voltage and the phase difference is zero in the phase diagram.
In option D the value of current is lesser than voltage but there is a phase difference in the phase diagram.
Therefore, option C is the correct answer for this problem.

Note: A phase diagram is used to show the relationship between two or more sine waves having the same frequency. While applying Kirchhoff’s rule if we follow the flow of current then we are proceeding from higher potential to the lower potential of the source.