Question

An Electric iron is rated 1000W, 240V. Find the current drawn & resistance of the heating element.

Answer 1

To solve for the current drawn and the resistance of the heating element of the electric iron, we need to use some basic electrical formulas. Let's go through the calculations step by step.

### Given Data
- Power rating (\( P \)) = 1000 W
- Voltage (\( V \)) = 240 V

### Finding the Current Drawn

We can use the formula relating power, voltage, and current:

\[ P = V \times I \]

where:
- \( P \) is the power in watts (W),
- \( V \) is the voltage in volts (V),
- \( I \) is the current in amperes (A).

Rearrange the formula to solve for \( I \):

\[ I = \frac{P}{V} \]

Substitute the given values:

\[ I = \frac{1000\ \text{W}}{240\ \text{V}} \]

\[ I = 4.167\ \text{A} \]

So, the current drawn by the iron is approximately **4.167 amperes**.

### Finding the Resistance of the Heating Element

We can use Ohm's Law, which states:

\[ V = I \times R \]

where:
- \( R \) is the resistance in ohms (Ω).

Rearrange this formula to solve for \( R \):

\[ R = \frac{V}{I} \]

Substitute the values we have:

\[ R = \frac{240\ \text{V}}{4.167\ \text{A}} \]

\[ R \approx 57.6\ \text{Ω} \]

So, the resistance of the heating element is approximately **57.6 ohms**.

### Summary
- **Current Drawn**: 4.167 A
- **Resistance of Heating Element**: 57.6 Ω

Answer 2

To find the current drawn and the resistance of the heating element for the electric iron:

1. **Current Drawn (I):**

   Use the formula for power: \( P = VI \)

   Rearranging for current: \( I = \frac{P}{V} \)

   Given: \( P = 1000 \text{ W} \) and \( V = 240 \text{ V} \)

   So, \( I = \frac{1000 \text{ W}}{240 \text{ V}} = 4.17 \text{ A} \)

2. **Resistance (R):**

   Use Ohm's law: \( R = \frac{V}{I} \)

   Given: \( V = 240 \text{ V} \) and \( I = 4.17 \text{ A} \)

   So, \( R = \frac{240 \text{ V}}{4.17 \text{ A}} \approx 57.6 \text{ }\Omega \)

The current drawn is approximately 4.17 A, and the resistance of the heating element is approximately 57.6 Ω.