Step 1: Understanding the Concept:
This problem deals with basic DC circuit analysis using Ohm's Law and the concept of Electromotive Force (EMF).
Electromotive Force is not a force in the mechanical sense; it is the energy provided per unit charge by a source (like a battery) to move charges around a complete circuit loop.
Ohm's Law provides the empirical relationship between the voltage across a conductor and the resulting electric current through it, given a constant resistance.
In a simple single-loop circuit, the total EMF drives the current through the total equivalent resistance of the path.
Step 2: Key Formula or Approach:
According to Ohm's Law (or Kirchhoff's Loop Rule for a single loop):
\[ I = \frac{E}{R_{\text{total}}} \]
Where:
\( I \) is the steady-state current in Amperes (A).
\( E \) is the total Electromotive Force in Volts (V).
\( R_{\text{total}} \) is the sum of all resistances in the loop, including internal resistance if mentioned.
Step 3: Detailed Explanation:
Given Parameters:
Total EMF of the source = 10 V.
Total loop resistance = \( 5 \Omega \).
The question implies that \( 5 \Omega \) is the aggregate resistance (the sum of any external resistors and the internal resistance of the battery).
Substituting the values into the algebraic relationship:
\[ I = \frac{10 \text{ V}}{5 \Omega} \]
\[ I = 2 \text{ A} \]
This means that every second, 2 Coulombs of electric charge pass through any given cross-section of the wire in the loop.
Since it is a single closed loop, the current is uniform throughout the entire circuit.
No charge is "consumed" as it flows; the energy provided by the EMF is simply dissipated as heat in the resistance according to Joule's heating law (\( P = I^{2}R \)).
Step 4: Final Answer:
The electric current flowing through the circuit loop is 2 A.