If the voltage to a circuit is doubled and the resistance is increased to three times the original value, what will be the final current?

If the voltage to a circuit is doubled and the resistance is increased to three times the original value, what will be the final current?

• A. 1/3 the original current
• B. 2/3 the original current
• C. 3 times the original current
• D. None of the above

Answer: (B)

To understand why the current in the circuit would be \( \frac{2}{3} \) of the original current when the voltage is doubled and the resistance is increased to three times its original value, we can apply Ohm's law. Ohm's law states that current (I) is equal to voltage (V) divided by resistance (R), formulated as \( I = \frac{V}{R} \). Let's denote the original voltage as \( V \) and the original resistance as \( R \). Therefore, the original current can be expressed as: \[ I_{\text{original}} = \frac{V}{R} \] When the voltage is doubled, it becomes \( 2V \). Meanwhile, the resistance is increased to three times its original value, making it \( 3R \). Substituting these new values into Ohm's law gives the new current: \[ I_{\text{new}} = \frac{2V}{3R} \] Now, we can compare the new current to the original current: \[ I_{\text{new}} = \frac{2}{3} \cdot \frac{V}{R} = \frac{2}{3} \

To understand why the current in the circuit would be ( \frac{2}{3} ) of the original current when the voltage is doubled and the resistance is increased to three times its original value, we can apply Ohm's law.

Ohm's law states that current (I) is equal to voltage (V) divided by resistance (R), formulated as ( I = \frac{V}{R} ).

Let's denote the original voltage as ( V ) and the original resistance as ( R ). Therefore, the original current can be expressed as:

[ I_{\text{original}} = \frac{V}{R} ]

When the voltage is doubled, it becomes ( 2V ). Meanwhile, the resistance is increased to three times its original value, making it ( 3R ). Substituting these new values into Ohm's law gives the new current:

[ I_{\text{new}} = \frac{2V}{3R} ]

Now, we can compare the new current to the original current:

[ I_{\text{new}} = \frac{2}{3} \cdot \frac{V}{R} = \frac{2}{3} \