For an electrical project in a residence, the electrical engineer had resistors resistances equal to 300 and 100 to assemble a circuit of an electrical appliance that must be connected 12
hours per day. In order to obtain a chain for optimal operation, he came to the conclusion that the best
configuration would be the one shown in the figure below (consider the middle thread as ideal). Knowing that
kWh value is $ 0.35, what is the amount that must be paid for energy consumption over 1 month (30 days)?

$ 0.077

Explanation:

We'll begin by calculating the equivalent resistance of the three 300 Ω resistors connected in series. This can be obtained as follow:

Equivalent resistance of the three 300 Ω (R₃₀₀) = 300 + 300 + 300

= 900 Ω

Next, we shall determine the equivalent resistance of the two 100 Ω resistors connected in series. This can be obtained as follow:

Equivalent resistance of the two 100 Ω (R₁₀₀) = 100 + 100

= 200 Ω

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

Equivalent resistance of the three 300 Ω (R₃₀₀) = 900 Ω

Equivalent resistance of the two 100 Ω (R₁₀₀) = 200 Ω

Equivalent Resistance (R) =?

R = R₃₀₀ × R₁₀₀ / R₃₀₀ + R₁₀₀ (since they are in parallel connections)

R = 900 × 200 / 900 + 200

R = 163.64 Ω

Next, we shall determine the energy in KWh. This can be obtained as follow:

Voltage (V) = 10 V

Resistance (R) = 163.64 Ω

Time (t) for 30 days = 12 × 30 = 360 h

Energy (E) =?

E = V²t / R

E = 10² × 360 / 163.64

E = 100 × 360 / 163.64

E = 36000 / 163.64

E = 220 Wh

Divide by 1000 to express in KWh

E = 220 Wh / 1000 = 0.22 KWh

Finally, we shall determine the amount paid for 1 month (30 days). This can be obtained as follow:

Cost per KWh = $ 0.35

Energy (E) = 0.22 KWh

Cost for 30 days =?

Cost for 30 days = Energy × Cost per KWh

Cost for 30 days = 0.22 × 0.35

Cost for 30 days = $ 0.077

Therefore, the amount that must be paid for 1 month is $ 0.077