College

Two resistors are wired in series with a 123 V battery. The first resistor has a resistance of 6 kΩ, and the second resistor has a resistance of 1 kΩ. What is the current in amperes that is leaving the positive terminal of the battery?

Answer :

Final answer:

The current in amperes that is leaving the positive terminal of the battery is approximately 0.0176 A.

Explanation:

In this question, we have two resistors wired in series with a 123 V battery. The first resistor has a resistance of 6 kQ (6000 ohms) and the second resistor has a resistance of 1 KQ (1000 ohms).

To find the current leaving the positive terminal of the battery, we need to calculate the total resistance of the circuit. In a series circuit, the total resistance is the sum of the individual resistances.

Total resistance (Rtotal) = R1 + R2

Substituting the given values:

Rtotal = 6000 ohms + 1000 ohms = 7000 ohms

Now, we can use Ohm's Law to calculate the current (I) in the circuit:

I = V / Rtotal

Substituting the given voltage:

I = 123 V / 7000 ohms

Calculating the value:

I ≈ 0.0176 A

Therefore, the current leaving the positive terminal of the battery is approximately 0.0176 amperes.

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