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A 6.00-Ω resistor and a 3.00-Ω resistor are connected in series across a 9.00-V battery.

1. What is the current [tex]i[/tex] through the circuit?
2. What is the voltage drop [tex]V[/tex] across the 6.00-Ω resistor?

Answer :

Final answer:

The current flowing through the 6.00-9 resistor is approximately 0.999 A, and the voltage drop across it is approximately 5.994 V.

Explanation:

In this series circuit, we have two resistors connected in series across a 9.00-V battery. The resistors have resistances of 6.00-92 and 3.00-12.

Since the resistors are connected in series, the same current flows through both resistors. Let's calculate the total resistance of the circuit:

Total resistance (Rtotal) = Resistance of the first resistor + Resistance of the second resistor

Rtotal = 6.00-92 + 3.00-12

Rtotal = 9.00-04

Now, we can use Ohm's Law to calculate the current flowing through the circuit:

Current (i) = Voltage (V) / Total resistance (Rtotal)

i = 9.00 / 9.00-04

i ≈ 0.999 A

The current flowing through the 6.00-9 resistor is approximately 0.999 A.

To calculate the voltage drop across the 6.00-9 resistor, we can use Ohm's Law:

Voltage drop (V) = Current (i) * Resistance (R)

V ≈ 0.999 * 6.00

V ≈ 5.994 V

The voltage drop across the 6.00-9 resistor is approximately 5.994 V.

Learn more about current and voltage in a series circuit here:

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