If the primary side of a transformer is connected with 230 V, 50 Hz A.C supply and the ratio of number of turns of primary to the secondary winding is 10 : 1, load resistance at secondary coil is 46 Ω then power output of the secondary windings is

The correct option is 11.5 watt.

Step-by-Step Explanation:

From the problem statement and the circuit schematic shown in the image, we can identify the following parameters:
- Primary voltage ($V_1$) = 230 V
- Primary to secondary turn ratio ($\frac{N_1}{N_2}$) = $\frac{10}{1}$
- Load resistance at the secondary winding ($R_L$) = 46 Ω

Step 1: Calculate the secondary voltage ($V_2$)
According to the transformer turns ratio formula shown in the image:

$\frac{V_2}{V_1}=\frac{N_2}{N_1}$

Rearranging the formula to solve for $V_2$:

$V_2=V_1\times \left(\frac{N_2}{N_1}\right)$

Substitute the given values:

$V_2=230\times \frac{1}{10}=23\text{V}$
This matches the 23 V potential difference labeled across the secondary winding in the schematic diagram.

Step 2: Calculate the power output at the secondary winding ($P_L$)
The power dissipated by the load resistor connected to the secondary coil is given by:

$P_L=\frac{{V_2}^2}{R_L}$

Substitute $V_2=23\text{V}$ and $R_L=46\Omega$ into the equation:

$P_L=\frac{{\left(23\right)}^2}{46}$

$P_L=\frac{529}{46}=11.5\text{W}$

Therefore, the power output of the secondary windings is 11.5 watt.