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A 6.00-Ω resistor and a 5.00-Ω resistor are connected in series across a 7.00-V battery.

What is the current through the 6.00-Ω resistor?

What is the voltage drop across the 6.00-Ω resistor?

Answer :

The current through the 6.00-ω resistor is 0.636 A and the voltage drop across it is 3.82 V. It can be calculated using Ohm's Law.

The current through and the voltage drop across the 6.00-ω resistor can be found by using Ohm's Law and the rules for series circuits.

First, we need to find the total resistance of the circuit. Since the resistors are in series, the total resistance is the sum of the individual resistances:

[tex]R_tot = R_1 + R_2 = 6.00 + 5.00 = 11.00[/tex] ω

Next, we can use Ohm's Law to find the current through the circuit:

I = V/R = 7.00/11.00 = 0.636 A

Since the resistors are in series, the current through each resistor is the same. So the current through the 6.00-ω resistor is also 0.636 A. Finally, we can use Ohm's Law again to find the voltage drop across the 6.00-ω resistor:

[tex]V = I*R = 0.636*6.00 = 3.82 V[/tex]

So the current through the 6.00-ω resistor is 0.636 A and the voltage drop across it is 3.82 V.

Here you can learn more about Ohm's Law https://brainly.com/question/1247379

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