What happens to the current in a circuit if the emf of the battery is doubled while keeping the resistance constant?

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Multiple Choice

What happens to the current in a circuit if the emf of the battery is doubled while keeping the resistance constant?

Explanation:
When the electromotive force (emf) of a battery is doubled while keeping the resistance constant, the current in the circuit increases. This relationship is described 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 and inversely proportional to the resistance (R). Mathematically, this is expressed as: I = V / R If the original emf (voltage) is V and it is doubled to 2V, while the resistance remains constant at R, the new current can be calculated as: I_new = 2V / R This shows that the new current is indeed double the original current, confirming that when the voltage increases while resistance is unchanged, the current will also increase proportionately. Therefore, if the voltage is doubled, the current in the circuit will be doubled as well.

When the electromotive force (emf) of a battery is doubled while keeping the resistance constant, the current in the circuit increases. This relationship is described 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 and inversely proportional to the resistance (R). Mathematically, this is expressed as:

I = V / R

If the original emf (voltage) is V and it is doubled to 2V, while the resistance remains constant at R, the new current can be calculated as:

I_new = 2V / R

This shows that the new current is indeed double the original current, confirming that when the voltage increases while resistance is unchanged, the current will also increase proportionately. Therefore, if the voltage is doubled, the current in the circuit will be doubled as well.

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