(a) \(\text{The current } (I) \text{ is calculated using Ohm's Law: } I = \frac{V}{R}.\)
For the 2000 W heater: \(\text{Resistance } R = 25\,\Omega\), \(\text{Voltage } V = 220\,\text{V}\): \[ I_{\text{heater}} = \frac{220}{25} = 8.8\,\text{A} \]
For the 100 W bulb: \(\text{Resistance } R = 500\,\Omega\), \(\text{Voltage } V = 220\,\text{V}\): \[ I_{\text{bulb}} = \frac{220}{500} = 0.44\,\text{A} \]
Since \(8.8\,\text{A} > 0.44\,\text{A}\), the 2000 W heater can carry a larger current.
\(I_{\text{bulb}} = \frac{220}{500} = 0.44\,\text{A}\)
Answer (a): The 2000 W heater carries a much larger current (8.8 A) compared to the 100 W bulb (0.44 A).
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