A bulb is rated \((100\mathrm{W},110\mathrm{V})\). It is operated by current of 1.0 A supplied by a step down transforme…

The power output of the transformer is the power consumed by the bulb, which is 100 W. The efficiency of the transformer is given as 0.9.

First, we calculate the power input to the transformer using the efficiency formula:

\(\eta = \frac{P_{out}}{P_{in}}\)

where \(\eta\) is the efficiency, \(P_{out}\) is the output power, and \(P_{in}\) is the input power.

Rearranging for \(P_{in}\):

\(P_{in} = \frac{P_{out}}{\eta}\)

Substituting the given values:

\(P_{in} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\)

Next, we use the input power and input voltage to find the input current. The input voltage is given as 220 V.

Using the power formula:

\(P_{in} = V_{in} \times I_{in}\)

Rearranging for \(I_{in}\):

\(I_{in} = \frac{P_{in}}{V_{in}}\)

Substituting the values:

\(I_{in} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\)

Thus, the input current drawn from the mains is approximately \(\frac{5}{9} \text{ A}\).
Identify the power consumed by the bulb: \(P_{out} = 100 \text{ W}\).

Calculate the input power using the efficiency: \(P_{in} = \frac{P_{out}}{\eta} = \frac{100 \text{ W}}{0.9} = 111.11 \text{ W}\).

Identify the input voltage: \(V_{in}\) = \(220 \text{ V}\).

Use the power formula to find the input current: \(P_{in} = V_{in} \times I_{in}\).

Rearrange and solve for \(I_{in}: I_{in} = \frac{P_{in}}{V_{in}} = \frac{111.11 \text{ W}}{220 \text{ V}} = 0.505 \text{ A}\).

final answer: \(\frac{5}{9} \text{ A}\)