Question

Connected to a battery that offers a d.d.p. of 3.0Volts, we have the four lamps, all identical with resistances equal to 2.0 Ohms. The association appears in the figure. Note that the switch is open, so it doesn't pass current in the circuit (it's like an off switch). When the key closes, the lamps come on. Determining the Req of the association and the total current flowing through the circuit, we will find:

Answer 1

Answer:

Option D. 5Ω e 0.6 A

Explanation:

1. Determination of the equivalent resistance.

We'll begin by calculating the equivalent resistance of resistor B and resistor C. This can be obtained as follow:

Resistor B (R₆) = 2 Ω

Resistor C (R꜀) = 2 Ω

Equivalent Resistance (R₆꜀) =?

R₆꜀ = R₆ × R꜀ / R₆ + R꜀ (parallel connection)

R₆꜀ = 2 × 2 / 2 + 2

R₆꜀ = 4 / 4

R₆꜀ = 1 Ω

Finally, we shall determine the equivalent resistance of the circuit. This can be obtained as follow:

Resistor A (Rₐ) = 2 Ω

Equivalent resistance of resistor B and C (R₆꜀) = 1 Ω

Resistor D (Rₔ) = 2 Ω

Equivalent Resistance (R) =?

R = Rₐ + R₆꜀ + Rₔ (series connection)

R = 2 + 1 + 2

R = 5 Ω

Thus, the equivalent resistance of the circuit is 5 Ω

2. Determination of the total current in the circuit.

Potential difference (V) = 3 V

Equivalent resistance (R) = 5 Ω

Total current (I) =?

V = IR

3 = I × 5

Divide both side by 5

I = 3 / 5

I = 0.6 A

Thus, the total current in the circuit is 0.6 A

Summary:

Equivalent resistance = 5 Ω

Current = 0.6 A

Option D gives the current answer to the question.