Question

The ratio of the number of turns in the primary to the secondary of an ideal transformer is \(1:5\). If \(5\) kW power at \(200\) V is supplied to the primary, find
(i) current in the primary, and
(ii) output voltage.

Hint:

For an ideal transformer: \[ P_p=P_s \] \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] \[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \] Voltage increases in the same ratio as the turns, while current decreases in the inverse ratio.

Answer 1

Concept: For an ideal transformer: \[ \text{Input Power} = \text{Output Power} \] and \[ \frac{V_s}{V_p} = \frac{N_s}{N_p}. \] The turns ratio directly determines the voltage ratio.

Step 1: Write the given data. Input power \[ P_p=5\text{ kW} =5000\text{ W} \] Input voltage \[ V_p=200\text{ V} \] Turns ratio \[ N_p:N_s=1:5 \] Hence, \[ \frac{N_s}{N_p}=5. \]

Step 2: Calculate the primary current. Using \[ P_p=V_pI_p, \] we get \[ 5000=200\times I_p. \] Therefore, \[ I_p=\frac{5000}{200}. \] \[ \boxed{ I_p=25\text{ A} } \]

Step 3: Calculate the output voltage. For an ideal transformer, \[ \frac{V_s}{V_p} = \frac{N_s}{N_p}. \] Substituting the values, \[ \frac{V_s}{200} = 5. \] Therefore, \[ V_s = 5\times200. \] \[ \boxed{ V_s=1000\text{ V} } \]

Step 4: Verification using power conservation. Since the transformer is ideal, \[ P_s=P_p=5000\text{ W}. \] Thus, \[ I_s = \frac{5000}{1000} = 5\text{ A}. \] This satisfies \[ \frac{I_s}{I_p} = \frac{5}{25} = \frac{1}{5} = \frac{N_p}{N_s}. \] Hence the result is correct. Final Answers: \[ \boxed{ I_p=25\text{ A} } \] \[ \boxed{ V_s=1000\text{ V} } \]