Question

Three resistors, 2.50 kΩ, 2.20 kΩ and 3.30 kΩ, are connected in series. A power supply is connected in the circuit, and applies a voltage of 10.00 V across all three. Calculate the current in the circuit. Enter results in terms of milliamps (mA). (Note 1 mA = 0.001 A)

Answer 1

To calculate the current in the circuit, sum the resistances (8.00 kΩ), then apply Ohm's Law (I = V / R). Divide the voltage (10.00 V) by the total resistance to get the current in amperes, and multiply by 1,000 to convert to milliamps, resulting in a current of 1.25 mA.

Step-by-step explanation:

The student is asking about calculating the current in an electrical circuit where three resistors are connected in series. To find the current, we 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) of the conductor. The formula to calculate current is I = V / R. In a series circuit, the total resistance (Rtotal) is the sum of all individual resistances.

Steps to Solve the Problem:

1. Calculate the total resistance by adding the values of all three resistors together.
2. Apply Ohm's Law to find the current by dividing the voltage supplied by the power source by the total resistance calculated in the first step.
3. Convert the current from amperes to milliamps as the question requests.

So, Rtotal = 2.50 kΩ + 2.20 kΩ + 3.30 kΩ = 8.00 kΩ

Now, using Ohm's Law I = V / Rtotal, we find I = 10.00 V / 8.00 kΩ = 0.00125 A.

Since 1 A = 1,000 mA, the current in milliamps is I = 0.00125 A × 1,000 mA/A = 1.25 mA.