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------------------------------------------------ Two 121 V batteries are connected in parallel to two 6 Ω resistors in series. How much current flows through each battery?

A) 0.1250 A
B) 0.0625 A
C) 0.2500 A
D) 0.5000 A
E) 1.000 A

To find out how much current flows through each battery, we need to first calculate the total resistance of the circuit. In this case, we have two 6Ω resistors connected in series. When resistors are connected in series, their resistances add up.

So, the total resistance of the circuit is 6Ω + 6Ω = 12Ω.

Next, we need to use LOhm's Law to find the total current flowing through the circuit. Ohm's Law states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is the same for both batteries and is equal to 121V.

Using Ohm's Law, we can calculate the total current:

I = V / R
I = 121V / 12Ω
I ≈ 10.0833 A

Since the resistors are in series, the same current flows through both of them. Therefore, the current flowing through each battery is approximately 10.0833 A.

To answer the question, none of the options provided (a, b, c, d, e) match the calculated value. However, the closest option is 0.2500 A (c). But it is important to note that this option is not exactly accurate, as the calculated value is slightly different.

In summary, the current flowing through each battery in this circuit is approximately 10.0833 A.

Learn more about circuit from the given link

https://brainly.com/question/2969220

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