A series circuit is shown in the diagram below. What is the potential drop across R₁? (R₂ = 20 Ω, R₂ = 40 Ω, R₃ = 60 Ω, V = 60 V) (Ohm's law: IR)

Option B. 10 V.

Explanation:

The following data were obtained from the question:

Resistor 1 (R₁) = 20 Ω

Resistor 2 (R₂) = 40 Ω

Resistor 3 (R₃) = 60 Ω

Potential difference (V) = 60 V

Next, we shall determine the total resistance of the circuit.

The circuit is in series connection. Therefore the total resistance (R) can be obtained by adding up R₁, R₂ and R₃ together as shown below:

R = R₁ + R₂ + R₃

R = 20 + 40 + 60

R = 120 Ω

Therefore, the total resistance in the circuit is 120 Ω

Next, we shall determine the current flowing in the circuit.

Note: The same current will flow through each resistor since they are in series connection.

The current flowing in the circuit can be obtained as follow:

Potential difference (V) = 60 V

Total resistance (R) = 120 Ω

Current (I) =..?

V = IR

60 = I x 120

Divide both side by 120

I = 60/120

I = 0.5 A

Therefore, the current flowing in the circuit is 0.5 A

Now, we can calculate the potential drop across R₁ as follow:

Resistor 1 (R₁) = 20 Ω

Current (I) = 0.5 A

Potential difference cross R₁ (V₁) =.?

V₁ = IR₁

V₁ = 0.5 x 20

V₁ = 10 V

Therefore, the potential difference across R₁ is 10 V.

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