Question 20 QMED02 - Electrician-Refrigerating Engineer

The Correct Answer is C ### Explanation for Option C (Correct) This scenario is governed by Ohm's Law, which states that the current ($I$) flowing through a conductor between two points is directly proportional to the voltage ($V$) across the two points, provided the temperature and other physical conditions remain constant. The relationship is expressed as: $$I = \frac{V}{R}$$ Where $R$ is the constant resistance. **Original Current ($I_{original}$):** $$I_{original} = \frac{V_{original}}{R}$$ **New Current ($I_{new}$):** The resistance ($R$) is held constant, and the applied voltage is halved ($\frac{V_{original}}{2}$). $$I_{new} = \frac{V_{new}}{R} = \frac{(V_{original} / 2)}{R} = \frac{1}{2} \times \left( \frac{V_{original}}{R} \right)$$ Substituting the original current equation: $$I_{new} = \frac{1}{2} \times I_{original}$$ Therefore, if the voltage is halved while the resistance is constant, the current flow will also be halved (or divided by two). ### Explanation for Incorrect Options **A) double:** This would happen if the voltage were doubled (while resistance remained constant) or if the resistance were halved (while voltage remained constant). Since the voltage was halved, this option is incorrect. **B) remain the same:** This would only happen if both the voltage and the resistance were kept constant, or if both were changed proportionally (e.g., if voltage were halved and resistance were also halved). Since the voltage was halved and the resistance was constant, the current must change, making this option incorrect. **D) be divided by four:** This relationship suggests a squared dependency, such as if the voltage were related to power ($P = V^2/R$) or if the voltage were halved and the resistance were doubled. Since current is linearly proportional to voltage ($I \propto V$), halving the voltage results in halving the current, not dividing it by four.