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Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is :

a.
1.33 A

b.
2.66 A

c.
12 A

Sagot :

AL2006

In a series circuit . . .

-- The total resistance is the sum of the individual resistors.

-- The current is the same at every point in the circuit.

The total resistance in this circuit is (3Ω + 6Ω )  =  9Ω

The current at every point is (V/R) = (12v / 9Ω ) = 1.33 A .

Pick choice (a).

Lanuel

The current flowing through the 6 Ohms resistor is: a.  1.33 A

Given the following data:

  • Resistor A = 3 Ohms.
  • Resistor B = 6 Ohms.
  • Voltage = 12 Volts.

To find the current flowing through the 6 Ohms resistor:

First of all, we would determine the total equivalent resistance of the resistors connected in series.

[tex]T_eq = Resistor \; A + Resistor\; B\\\\T_eq = 3 + 6[/tex]

Total equivalent resistance = 9 Ohms

Note: The current flowing through both resistors are the same since they are connected in series.

From Ohm's law, we have:

[tex]Current = \frac{Voltage}{Resistor}\\\\Current = \frac{12}{9}[/tex]

Current = 1.33 Amperes

Therefore, the current flowing through the 6 Ohms resistor is 1.33 Amperes.

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