To solve this problem, we can use Ohm's law, which states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) between them. The formula is I = V/R.

1. First, we need to find the total voltage in the circuit. The battery is being charged from a 15-V source, but the battery itself has a voltage of 12 V. Because the battery is being charged, its voltage is in the opposite direction of the source voltage. So, the total voltage is 15 V - 12 V = 3 V.

2. We know that the charging current should not exceed 2 A. So, we can set I = 2 A.

3. Now, we can use Ohm's law to find the total resistance needed to keep the current at 2 A with a voltage of 3 V. Rearranging the formula to solve for R gives us R = V/I = 3 V / 2 A = 1.5 Ω.

4. However, we know that the battery already has an internal resistance of 0.3 Ω. So, the resistance of the series resistor we need to add is 1.5 Ω - 0.3 Ω = 1.2 Ω.

So, the minimum resistance of a series resistor that will limit the current to 2 A is 1.2 Ω.

Question

To solve this problem, we can use Ohm's law, which states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) between them. The formula is I = V/R.

1. First, we need to find the total voltage in the circuit. The battery is being charged from a 15-V source, but the battery itself has a voltage of 12 V. Because the battery is being charged, its voltage is in the opposite direction of the source voltage. So, the total voltage is 15 V - 12 V = 3 V.

2. We know that the charging current should not exceed 2 A. So, we can set I = 2 A.

3. Now, we can use Ohm's law to find the total resistance needed to keep the current at 2 A with a voltage of 3 V. Rearranging the formula to solve for R gives us R = V/I = 3 V / 2 A = 1.5 Ω.

4. However, we know that the battery already has an internal resistance of 0.3 Ω. So, the resistance of the series resistor we need to add is 1.5 Ω - 0.3 Ω = 1.2 Ω.

So, the minimum resistance of a series resistor that will limit the current to 2 A is 1.2 Ω.

Knowee AI · Accepted Answer

To solve this problem, we can use Ohm's law, which states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) between them. The formula is I = V/R.

1. First, we need to find the total voltage in the circuit. The battery is being charged from a 15-V source, but the battery itself has a voltage of 12 V. Because the battery is being charged, its voltage is in the opposite direction of the source voltage. So, the total voltage is 15 V - 12 V = 3 V.

2. We know that the charging current should not exceed 2 A. So, we can set I = 2 A.

3. Now, we can use Ohm's law to find the total resistance needed to keep the current at 2 A with a voltage of 3 V. Rearranging the formula to solve for R gives us R = V/I = 3 V / 2 A = 1.5 Ω.

4. However, we know that the battery already has an internal resistance of 0.3 Ω. So, the resistance of the series resistor we need to add is 1.5 Ω - 0.3 Ω = 1.2 Ω.

So, the minimum resistance of a series resistor that will limit the current to 2 A is 1.2 Ω.

A 12-V battery with a 0 . 3 - 0 internal resistance is to be charged from a 15-V source. If the charging currentshould not exceed 2 A, what is the minimum resistance of a series resistor that will limit the current to thissafe value

Question

Solution

Similar Questions

Upgrade your grade with Knowee