Transformer

$P=VI\Rightarrow I=\dfrac{88}{11000}=8\times10^{-3}\ A=8$ mA.

For a transformer: $\frac{i_p}{i_s}=\frac{V_s}{V_p}=\frac{N_s}{N_p}$ For the bulb, $V_s=12\ V$; $i_s=\frac{60}{12}=5\ A$ $\Rightarrow\frac{i_p}{5}=\frac{12}{220}$ $i_p=\frac{60}{220}=\frac{3}{11}=0.27\ A$

For an ideal transformer, input power = output power: $220 \times i_p = 44$ $i_p = \frac{44}{220} = 0.2$ A

Efficiency $\eta = \frac{\text{output power}}{\text{input power}} = \frac{E_2I_2}{E_1I_1}$
Output power = $440\times2.0 = 880$ W
Input power = $\frac{880}{0.80} = 1100$ W
$I_1 = \frac{\text{input power}}{E_1} = \frac{1100}{220}$
$I_1 = 5.0$ A

For a transformer: $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$
Here $N_p$ = number of turns in the primary coil = 50 and $N_s$ = number of turns in the secondary coil = 1500.
The emf across the primary equals the rate of change of the flux linked with it: $V_p = \dfrac{d\phi}{dt} = \dfrac{d}{dt}(\phi_0 + 4t) = 4$ V
$\phi_0$ is a constant, so its derivative is zero.
$V_s = \dfrac{N_s}{N_p} \times V_p = \dfrac{1500}{50} \times 4$
$V_s = 30 \times 4 = 120$ V
So the output voltage across the secondary coil is 120 volts.

Efficiency of the transformer: $\eta = \dfrac{P_{output}}{P_{input}} \times 100$
Output power = power of the lamp = 100 W
Input power $= V_p I_p = 220 \times 0.5 = 110$ W
$\eta = \dfrac{100}{220 \times 0.5} \times 100 = \dfrac{100}{110} \times 100$
$\eta = 90.9$ percent
So the efficiency is approximately 90%.