Q.A step-up transformer has $200$ turns in its primary coil and $4000$ turns in its secondary coil. The primary is connected to a $220\ \text{V}$ AC supply, and the transformer delivers a current of $2\ \text{A}$ to the secondary. Assuming an ideal (100% efficient) transformer, calculate

Given: $N_p=200$, $N_s=4000$, $V_p=220\ \text{V}$, $I_s=2\ \text{A}$, ideal (100% efficient) transformer.

1. Secondary voltage.

   $V_s = V_p\cdot\frac{N_s}{N_p} = 220\times\frac{4000}{200} = 220\times20 = 4400\ \text{V}$

2. Power delivered to the secondary.

   $P_s = V_sI_s = 4400\times2 = 8800\ \text{W}$

3. Primary current. For an ideal transformer, input power equals output power, $P_p=P_s=8800\ \text{W}$, so $I_p = \frac{P_p}{V_p} = \frac{8800}{220} = 40\ \text{A}$ …

Apply the turns-ratio relation to find $V_s$, compute the secondary power $V_sI_s$, then use ideal power conse …

• Inverting the turns ratio (using $N_p/N_s$ instead of $N_s/N_p$) when finding $V_s$, giving a step-down result for what is stated to be a step-up transformer. …