Question 6A62

This question relies on **Ohm's Law**, which states that current ($I$) is directly proportional to voltage ($V$) and inversely proportional to resistance ($R$). The formula is $I = V/R$. Let's denote the original voltage, resistance, and current as $V_1$, $R_1$, and $I_1$ respectively. So, $I_1 = V_1 / R_1$. Now, consider the new conditions: * The voltage is doubled: $V_2 = 2 \times V_1$ * The resistance is tripled: $R_2 = 3 \times R_1$ Substitute these into Ohm's Law to find the new current ($I_2$): $I_2 = V_2 / R_2 = (2 \times V_1) / (3 \times R_1)$ This can be rewritten as: $I_2 = (2/3) \times (V_1 / R_1)$ Since $I_1 = V_1 / R_1$, we can substitute $I_1$ into the equation: $I_2 = (2/3) \times I_1$ Therefore, the final current will be 2/3 the original current. * **A) 1/3 the original current:** This would be true if the voltage remained the same while resistance tripled ($1V/3R = 1/3I$). * **C) 3 times the original current:** This would imply a much larger increase in voltage relative to resistance, or a decrease in resistance. * **D) None of the above:** Incorrect, as option B correctly describes the outcome.