Question

An ideal transformer with purely resistive load operates at 12 kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be ________ mΩ.

Answer 1

Correct Answer: 240

\(\frac{𝑉_𝑠}{𝑉_𝑝} = \frac{𝑁_𝑠}{𝑁_𝑝}\)
⟹ \(\frac{120}{12000} = \frac{𝑁_𝑠}{𝑁_𝑝}\) 
⟹ \(\frac{𝑁_𝑠}{𝑁_𝑝} = \frac{1}{100}\) − − − (𝑖) 
For an ideal transformer, input power = Output power 
And power is given by 𝑃 = 𝑖𝑉 
\(𝑖_𝑝𝑉_𝑝 = 𝑖_𝑠𝑉_𝑠 = 60000𝑊\) 
\(𝑖𝑝 = \frac{60000}{12000} = 5\) 
Now, \(𝑅_𝑝 = \frac{𝑉_𝑝}{𝑖_𝑝}\)\( = \frac{12000}{5} = 2400 Ω \)
\(𝑅_𝑠 =\frac{𝑉_𝑠}{𝑖_𝑠} = \frac{120}{60000/120} = 120 × 120/60000 = 120/500 = 0.240Ω = 240 𝑚Ω\)

Answer 2

Transformer Voltage Ratio and Current Calculation 

Step 1: Transformer Voltage Ratio

The voltage ratio in a transformer is equal to the turns ratio:

\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)

where \( V_s \) and \( V_p \) are the secondary and primary voltages, and \( N_s \) and \( N_p \) are the number of turns in the secondary and primary coils, respectively. Given \( V_p = 12 \, \text{kV} = 12000 \, \text{V} \) and \( V_s = 120 \, \text{V} \), we have:

\( \frac{120}{12000} = \frac{N_s}{N_p} \quad \Rightarrow \quad \frac{N_s}{N_p} = \frac{1}{100} \)

Step 2: Power Conservation

For an ideal transformer, the input power is equal to the output power:

\( P_{in} = P_{out} \)

Power is given by \( P = IV \), so:

\( I_p V_p = I_s V_s = 60 \, \text{kW} = 60000 \, \text{W} \)

Step 3: Primary Current

We can calculate the primary current (\( I_p \)):

\( I_p = \frac{60000}{V_p} = \frac{60000}{12000} = 5 \, \text{A} \)

Step 4: Secondary Current

Similarly, we can calculate the secondary current (\( I_s \)):

\( I_s = \frac{60000}{V_s} = \frac{60000}{120} = 500 \, \text{A} \)

Step 5: Secondary Resistance

The secondary resistance (\( R_s \)) can be calculated using Ohm’s law:

\( R_s = \frac{V_s}{I_s} = \frac{120}{500} = 0.24 \, \Omega = 240 \, \text{m}\Omega \)

Conclusion:

The value of the resistive load in the secondary circuit is \( 240 \, \text{m}\Omega \).