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Answer :
Final answer:
Applying Ohm's Law, the current through the 4.00-Ohm resistor is the same as that through the 8.00-Ohm resistor, which is .75A. Using Ohm's law again, the voltage drop across the 4.00-Ohm resistor is calculated to be 3 V.
Explanation:
Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.
The resistors being connected in series means the same current flows through both of them. Based on Ohm's law, given as V=IR, where V is the voltage, I is the current, and R is the resistance, you provided that the current i1 through the 8.00-Ω resistor is .75 A, and the voltage V1 drop across this resistor is 6V.
As the current in a series circuit is the same across all components, the current through the 4.00-Ω resistor i2 is also .75 A. To find the voltage drop V2 across the 4.00-Ω resistor, we can again use Ohm's law: V2 = i2* R2 which equals .75A * 4.00 Ω = 3 V. Hence, the current through the 4.00-Ω resistor is .75 A, and the voltage drop across it is 3 V.
Learn more about Ohm's Law here:
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