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A hi-fi set of power 40 W plays for 8 hours every day. Calculate the cost of playing the hi-fi for a month (31 days) if 1 kWh costs M0.80.

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2 parts
b

30 A30 \text{ A}

c

M3.33

Step-by-step solution

6 steps

Here are the solutions to your problems.

1.

Step 1: Calculate the daily energy consumption of the hi-fi set.
The power is 40 W40 \text{ W} and it plays for 8 hours8 \text{ hours} a day.

Edaily=P×tdailyE_{daily} = P \times t_{daily} Edaily=40 W×8 h=320 WhE_{daily} = 40 \text{ W} \times 8 \text{ h} = 320 \text{ Wh}

Step 2: Calculate the monthly energy consumption.
For a month of 31 days31 \text{ days}:

Emonthly=Edaily×31 daysE_{monthly} = E_{daily} \times 31 \text{ days} Emonthly=320 Wh/day×31 days=9920 WhE_{monthly} = 320 \text{ Wh/day} \times 31 \text{ days} = 9920 \text{ Wh}

Step 3: Convert the monthly energy consumption to kilowatt-hours (kWh).

Emonthly,kWh=9920 Wh1000 Wh/kWh=9.92 kWhE_{monthly, kWh} = \frac{9920 \text{ Wh}}{1000 \text{ Wh/kWh}} = 9.92 \text{ kWh}

Step 4: Calculate the total cost.
The cost is M0.80M0.80 per kWh.

Cost=Emonthly,kWh×Cost per kWh\text{Cost} = E_{monthly, kWh} \times \text{Cost per kWh} Cost=9.92 kWh×M0.80/kWh\text{Cost} = 9.92 \text{ kWh} \times M0.80/\text{kWh} Cost=M7.936\text{Cost} = M7.936

The cost of playing the hi-fi for a month is M7.94\boxed{M7.94}.

2.

Part a

The different cooking elements (oven and hot plates) are likely connected in parallel. This is because connecting them in parallel ensures that each element receives the full 240 V240 \text{ V} mains supply, allowing them to operate independently at their rated power. If they were in series, the voltage would divide, and they would not function correctly or efficiently.

Part b

30 A30 \text{ A}

  1. Calculate the total power of all cooking elements when they are on.

    Oven power Poven=3000 WP_{oven} = 3000 \text{ W}.
    Each hot plate power Pplate=1000 WP_{plate} = 1000 \text{ W}. There are two hot plates.

    Ptotal=Poven+2×PplateP_{total} = P_{oven} + 2 \times P_{plate} Ptotal=3000 W+2×1000 WP_{total} = 3000 \text{ W} + 2 \times 1000 \text{ W} Ptotal=3000 W+2000 W=5000 WP_{total} = 3000 \text{ W} + 2000 \text{ W} = 5000 \text{ W}

  2. Calculate the total current drawn by the stove.

    Using the formula P=V×IP = V \times I, we can find the current I=PVI = \frac{P}{V}.

    I=5000 W240 VI = \frac{5000 \text{ W}}{240 \text{ V}} I≈20.83 AI \approx 20.83 \text{ A}

  3. Determine a suitable fuse value.

    A fuse should have a rating slightly higher than the maximum operating current to prevent it from blowing unnecessarily, but low enough to protect the circuit from overcurrent. Standard fuse ratings often include 5 A5 \text{ A}, 13 A13 \text{ A}, 30 A30 \text{ A}. Since 20.83 A20.83 \text{ A} is the operating current, a 30 A30 \text{ A} fuse would be suitable.
    The value of a suitable fuse is 30 A\boxed{30 \text{ A}}.

Part cM3.33

  1. Identify the total power and convert the operating time to hours.

    Total power Ptotal=5000 WP_{total} = 5000 \text{ W} (from part b).
    Operating time t=50 minutest = 50 \text{ minutes}.

    t=5060 hours=56 hourst = \frac{50}{60} \text{ hours} = \frac{5}{6} \text{ hours}

  2. Calculate the energy consumed in kilowatt-hours (kWh).

    E=Ptotal×tE = P_{total} \times t E=5000 W×56 hE = 5000 \text{ W} \times \frac{5}{6} \text{ h} E=250006 Wh≈4166.67 WhE = \frac{25000}{6} \text{ Wh} \approx 4166.67 \text{ Wh} EkWh=4166.67 Wh1000 Wh/kWh≈4.1667 kWhE_{kWh} = \frac{4166.67 \text{ Wh}}{1000 \text{ Wh/kWh}} \approx 4.1667 \text{ kWh}

  3. Calculate the total cost.

    The cost is M0.80M0.80 per kWh.

    Cost=EkWh×Cost per kWh\text{Cost} = E_{kWh} \times \text{Cost per kWh} Cost=4.1667 kWh×M0.80/kWh\text{Cost} = 4.1667 \text{ kWh} \times M0.80/\text{kWh} Cost≈M3.33336\text{Cost} \approx M3.33336

    The cost of operating all three cooking elements for 50 minutes is M3.33\boxed{M3.33}.

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