Calculate the efficiency of the transformer using the provided power and current values.

The transformer lights up four 12 V, 24 W light bulbs.
Total output power ($P_{out}$) = Number of bulbs $\times$ Power per bulb

From , the transformer takes $0.55 \, \text{A}$ from a $240 \, \text{V}$ mains. This is the input power ($P_{in}$).

Efficiency ($\eta$) is the ratio of output power to input power, expressed as a percentage.

The efficiency of the transformer is $\boxed{\text{72.73\%}}$.

98.

The transformer supplies $12 \, \text{V}$ (output) from a $240 \, \text{V}$ mains (input voltage, $V_p$).
It takes $0.55 \, \text{A}$ from the mains (input current, $I_p$).

The power delivered to the transformer ($P_{in}$) is the product of the input voltage and input current.

The power delivered to the transformer is $\boxed{\text{132 W}}$.

99.

Resistance of the cable ($R$) = $0.5 \, \Omega$.
Electricity generated ($P_{gen}$) = $11 \, \text{kW} = 11000 \, \text{W}$.
Transmission voltage ($V$) = $120 \, \text{V}$.

The current ($I$) can be found using the formula $P = V I$.

The power wasted ($P_{wasted}$) due to the resistance of the cable is given by $P_{wasted} = I^2 R$.

To express in kilowatts:

The power wasted in the cable is $\boxed{\text{4.20 kW}}$.