A ` 1 k W` heater is meant to operate at `200 V`.
(a) What is the resistance?
(b) How much power will it consume if the line voltage drops to `100 V`?
( c ) How many units of electrical energy will it consume in a month `( of 30 days)` if it operates `10 h `daily at the specified voltage `( 200 V)`?

Explanation

Let's solve the problem step by step.

### Given Data:
- Power of the heater, \( P = 1 \text{ kW} = 1000 \text{ W} \)
- Operating voltage, \( V = 200 \text{ V} \)

### (a) Finding the Resistance

To find the resistance \( R \) of the heater, we can use the formula:
\[ R = \frac{V^2}{P} \]

Substituting the values:
\[ R = \frac{200^2}{1000} = \frac{40000}{1000} = 40 \, \Omega \]

### (b) Power Consumption at 100 V

When the line voltage drops to \( 100 \text{ V} \), the resistance remains the same. We can find the new power \( P' \) using the formula:
\[ P' = \frac{V'^2}{R} \]
where \( V' = 100 \text{ V} \).

Substituting the values:
\[ P' = \frac{100^2}{40} = \frac{10000}{40} = 250 \, \text{W} \]

### (c) Total Energy Consumption in a Month

To find the total units of electrical energy consumed in a month, we first calculate the daily energy consumption. The formula for energy in kilowatt-hours is:
\[ \text{Energy (kWh)} = \text{Power (kW)} \times \text{Time (h)} \]

For one day:
- Power at specified voltage (200 V) = \( 1 \text{ kW} \)
- Time = \( 10 \text{ h} \)

Calculating daily energy consumption:
\[ \text{Energy (kWh)} = 1 \times 10 = 10 \text{ kWh} \]

Now, for 30 days:
\[ \text{Total Energy} = 10 \text{ kWh/day} \times 30 \text{ days} = 300 \text{ kWh} \]

### Summary of Answers:
(a) Resistance = \( 40 \, \Omega \)
(b) Power at 100 V = \( 250 \, \text{W} \)
(c) Total energy consumption in a month = \( 300 \, \text{kWh} \)

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