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.
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