What is the total emf, when three cells of emfs 2 V, 2 V, 2 V are connected in parallel?

We'll start by defining the word "electromotive force" here (emf). The relationship between the terminal potential difference (V) and emf will then be recorded (E). When cells from the same emf are linked in parallel, we will then put down the key notion.

Given that the first cell's emf equals the second cell's emf, and the third cell's emf equals two V

The potential difference that forms in an open circuit between a battery's two terminals is known as EMF, or electromotive force. When there is no current flowing through it, the emf is the theoretical difference between the positive terminal anode and the negative terminal cathode. The energy that is transmitted from a cell or battery to the charge carried is tested by the emf. The joules of energy are split by the coulombs charge. The emf acts as the catalyst for the current to flow.
The terminal potential difference is provided by V = E - Ir when internal resistance is taken into account, where V stands for the terminal potential difference, E for emf, I for current, and r for internal resistance.

Typically, a cell's internal resistance is not taken into account because $\epsilon \gg$ Ir. Internal resistance varies from cell to cell in terms of value.

Internal cellular resistance is not mentioned in this instance either. Therefore, we shall disregard it and assume that the electromotive force is equal to the terminal potential difference between the terminals (emf).
The net equivalent emf of the system will be equal to the individual emf E when n number of cells, each with an emf of E, are linked in parallel with one another.
It is obvious that all three parallel-connected cells have equal emfs. As a result, the individual emf of the cell will be equal to the overall emf of the circuit.
As a result, the system's necessary total emf will be 2 V.

Hence, the correct answer is option C.