Answer:
Option B. 2 A.
Explanation:
The following data were obtained from the question:
Resistance 1 (R1) = 1 Ω
Resistance 2 (R2) = 2 Ω
Resistance 3 (R3) = 3 Ω
Potential difference (V) = 12 V.
Current through the 3 Ω (I3) =?
Next, we shall determine the total resistance of the circuit.
This can be obtained as follow:
Resistance 1 (R1) = 1 Ω
Resistance 2 (R2) = 2 Ω
Resistance 3 (R3) = 3 Ω
Total resistance (R) =.?
R = R1 + R2 + R3
R = 1 + 2 + 3
R = 6 Ω
Next, we shall determine the current flowing in the circuit.
This can be obtained as follow:
From ohm's law,
Voltage (V) = current (I) x resistance (R)
V = IR
Potential difference (V) = 12 V.
Resistance (R) = 6 Ω
Current (I) =?
V = IR
12 = I x 6
Divide both side by 6
I = 12/6
I = 2 A.
Since the circuit is in series connection, therefore, the same current will flow through each resistor.
Therefore, the current through the the 3 Ω load is 2 A.
