1. Concept:Electrical charge is defined as the product of current and time (\( Q = I \times t \)). By fundamental definition, the electric potential difference (Voltage) is the work done or energy transferred per unit charge (\( V = \frac{W}{Q} \)). Therefore, the energy transferred by each coulomb of charge is equal to the voltage.
2. Formula:$$ Q = I t \quad \text{and} \quad W = V \times Q $$
3. Solution:- Charge in one second:
\( Q = I \times t = (0.25\text{ A}) \times (1\text{ s}) = 0.25\text{ C} \).
- Energy per coulomb:
By definition of voltage, \( V = \frac{W}{Q} \implies W = V \times Q \).
For each coulomb (\( Q = 1\text{ C} \)):
\( W = 2.2\text{ V} \times 1\text{ C} = 2.2\text{ J} \).
- Historical Exam Key Note: In MDCAT 2018, the examiner mistakenly calculated total energy in one second (\( E_{\text{total}} = V I t = 0.55\text{ J} \)) and marked Option D on the official key. However, the question specifically asks for energy transferred by the passage of each coulomb of charge, which is mathematically and physically \( 2.2\text{ J} \) (Option B).
4. Why other options are incorrect:- Option A: Incorrect charge calculation (\( 2.5\text{ C} \) instead of \( 0.25\text{ C} \)).
- Option B: Correct. Correctly identifies the charge passed in one second (\( 0.25\text{ C} \)) and the energy transferred per coulomb (\( 2.2\text{ J} \)).
- Option C: Lists potential difference in volts (\( 2.2\text{ V} \)), but energy must be expressed in joules (\( \text{J} \)).
- Option D: Common exam trap. \( 0.55\text{ J} \) is the total energy transferred by all charges in one second (\( E_{\text{total}} = V I t = 0.55\text{ J} \)), not the energy transferred per single coulomb.
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