Question

An ideal transformer is designed to convert 50 V into 250 V. It draws 200 W power from an AC source whose instantaneous voltage is given by \( v_i = 20 \sin(100\pi t) \, \text{V} \).

Find:

• 1. RMS value of input current.
• 2. Expression for instantaneous output voltage.
• 3. Expression for instantaneous output current.

Answer 1

(I) RMS Value of Input Current:

Provided data:

• Output voltage \( V_s = 250 \, \text{V} \)
• Input voltage \( V_p = 50 \, \text{V} \)
• Power from AC source \( P = 200 \, \text{W} \)
• Input instantaneous voltage \( v_i = 20 \sin(100\pi t) \, \text{V} \)

1. RMS value of input voltage: The input instantaneous voltage is given as \( v_i = V_p \sin(\omega t) \), where \( V_p = 20 \, \text{V} \) is the peak voltage. The RMS value \( V_{\text{rms}} \) is calculated as:

\[ V_{\text{rms}} = \frac{V_p}{\sqrt{2}} = \frac{20}{\sqrt{2}} \approx 14.14 \, \text{V} \]

2. RMS value of input current: The power \( P \) supplied by the AC source is related to the RMS voltage and current by \( P = V_{\text{rms}} I_{\text{rms}} \cos \phi \). For an ideal transformer, the power factor \( \cos \phi = 1 \). Therefore, the RMS input current \( I_{\text{rms}} \) is:

\[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{200}{14.14} \approx 14.14 \, \text{A} \]

The RMS value of the input current is approximately \( 14.14 \, \text{A} \).

(II) Expression for Instantaneous Output Voltage:

For an ideal transformer, the voltage ratio equals the turns ratio: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Given \( V_p = 50 \, \text{V} \) and \( V_s = 250 \, \text{V} \), the voltage is stepped up by a factor of 5. Thus, the instantaneous output voltage \( v_s \) is:

\[ v_s = 5 v_i = 5 \times 20 \sin(100\pi t) = 100 \sin(100\pi t) \, \text{V} \]

The expression for the instantaneous output voltage is \( v_s = 100 \sin(100\pi t) \, \text{V} \).

(III) Expression for Instantaneous Output Current:

The current relationship in an ideal transformer is \( \frac{I_s}{I_p} = \frac{V_p}{V_s} \). With \( V_p = 50 \, \text{V} \) and \( V_s = 250 \, \text{V} \), the secondary current is reduced by a factor of 5 compared to the primary current. The instantaneous output current \( i_s \) is related to the input current \( i_p \) by \( i_s = \frac{I_p}{5} \). By Ohm's law, the input current \( i_p \) is:

\[ i_p = \frac{v_i}{R} = \frac{20 \sin(100\pi t)}{400} = 0.05 \sin(100\pi t) \, \text{A} \]

Therefore, the instantaneous output current \( i_s \) is:

\[ i_s = \frac{0.05 \sin(100\pi t)}{5} = 0.01 \sin(100\pi t) \, \text{A} \]

The expression for the instantaneous output current is \( i_s = 0.01 \sin(100\pi t) \, \text{A} \).