If the voltage in a circuit is doubled, what happens to the voltage drop required across resistors?

When the voltage in a circuit is doubled, the voltage drop across resistors in that circuit follows Ohm's Law, which states that the voltage drop (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by its resistance (R), represented by the formula V = I * R.

If the voltage supplied to the circuit is increased (for instance, from V to 2V) while maintaining the same resistance, the overall current in the circuit would also increase, assuming that the circuit components are capable of carrying that additional current. The voltage drop across each resistor would then also increase proportionally.

In particular, if we're considering resistors in series, the total voltage provided by the source must equal the sum of the voltage drops across each resistor in that series. Doubling the voltage consequently requires that each voltage drop across the resistors also changes accordingly to maintain the balance dictated by Ohm's Law. Therefore, when the source voltage is doubled, the necessary voltage drop required across each resistor will also double, leading to the conclusion that the correct answer indicates that the voltage drop across resistors indeed doubles when the overall voltage is doubled.