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.
Learn more about calculating current in a series circuit here:
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