A transformer has an input Voltage of 400kV. If the Output Voltage is 100kV and number of turns in the out puts is 5000, with Current output of 20A determine the i. Number of turns in the output. ii. The current in the input.

The relationship between voltage and turns in an ideal transformer is given by:

$$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$

Rearrange the formula to solve for $N_s$:

$$N_s = N_p \left( \frac{V_s}{V_p} \right)$$

$$N_s = 5000 \text{ turns} \times \left( \frac{100 \text{ kV}}{400 \text{ kV}} \right)$$ $$N_s = 5000 \text{ turns} \times \left( \frac{1}{4} \right)$$ $$N_s = 1250 \text{ turns}$$

The number of turns in the output is $\boxed{\text{1250 turns}}$.

For an ideal transformer, the power in the primary coil equals the power in the secondary coil ($V_p I_p = V_s I_s$). This leads to the relationship:

$$\frac{V_p}{V_s} = \frac{I_s}{I_p}$$

Rearrange the formula to solve for $I_p$:

$$I_p = I_s \left( \frac{V_s}{V_p} \right)$$

$$I_p = 20 \text{ A} \times \left( \frac{100 \text{ kV}}{400 \text{ kV}} \right)$$ $$I_p = 20 \text{ A} \times \left( \frac{1}{4} \right)$$ $$I_p = 5 \text{ A}$$

The current in the input is $\boxed{\text{5 A}}$.