Below image, please give me the answer quick.

Option A. ⅓V

Explanation:

We'll begin by calculating the total resistance in the circuit. This is illustrated below:

R1 = 1Ω

R2 = 2Ω

R3 = 3Ω

RT =..?

Since all three resistor are in series connections, the total resistance can obtain as follow:

RT = R1 + R2 + R3

RT = 1 + 2 + 3

RT = 6Ω

Next we shall determine the current flowing through the circuit.

Voltage (V) = 2V

Resistor (R) = 6Ω

Current (I) =..?

V = IR

2 = I x 6

Divide both side by 6

I = 2/6

I = ⅓A

Finally, we can obtain the voltage across the 1Ω resistor as follow:

Resistor (R) = 1Ω

Current (I) = ⅓A

Voltage (V) =?

V = IR

V = ⅓ × 1

V = ⅓V

Therefore, the potential difference across the 1Ω resistor is ⅓V

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