A TV tube requires a 20.0 -kV-amplitude power supply. (a) What is the turns ratio of the transformer that raises the 170 -V-amplitude household voltage to \(20.0 \mathrm{kV} ?\) (b) If the tube draws 82 W of power, find the currents in the primary and secondary windings. Assume an ideal transformer.

The turns ratio of a transformer is the ratio of the number of turns in the primary winding to the number of turns in the secondary winding, and it is related to the voltage ratio between the primary and secondary windings. The formula for the turns ratio, in terms of voltages, is: $$ \frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}} $$ Where \(N_{1}\) and \(N_{2}\) are the number of turns in the primary and secondary windings, respectively, and \(V_{1}\) and \(V_{2}\) are the voltages across the primary and secondary windings, respectively.

Given the household voltage of 170 V and the required voltage of \(20.0kV\) for the TV tube, we can find the turns ratio by inserting these values into the formula: $$ \frac{N_{1}}{N_{2}} = \frac{170}{20,000} $$ Now, calculate the turns ratio: $$ \frac{N_{1}}{N_{2}} = 0.0085 $$ The turns ratio of the transformer is \(0.0085\). #b) Finding the currents in the primary and secondary windings#

When dealing with ideal transformers, the input power equals the output power: $$ P_{1} = P_{2} $$ Where \(P_{1}\) is the input power and \(P_{2}\) is the output power. Since power (\(P\)) is equal to the product of voltage (\(V\)) and current (\(I\)), we can write this equation in terms of current: $$ V_{1}I_{1} = V_{2}I_{2} $$ We were given that the tube draws 82 W of power, and we want to find \(I_{1}\) and \(I_{2}\).

We are given the output power and voltage, so we can find the current in the secondary winding, \(I_{2}\): $$ P_{2} = V_{2}I_{2} $$ Rearrange for \(I_{2}\), and substitute the given values: $$ I_{2} = \frac{P_{2}}{V_{2}} = \frac{82}{20,000} $$ Now, calculate the current in the secondary winding: $$ I_{2} = 0.0041\,\mathrm{A} $$ The current in the secondary winding is \(0.0041\,\mathrm{A}\).

We can use the turns ratio and the current in the secondary winding to find the current in the primary winding. Recall the turns ratio formula in terms of current: $$ \frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}} $$ We found the turns ratio in step 2, and the current in the secondary winding in step 4. Rearrange for \(I_{1}\), and substitute these values in the equation: $$ I_{1} = \frac{I_{2}}{\frac{N_{1}}{N_{2}}} = \frac{0.0041}{0.0085} $$ Now, calculate the current in the primary winding: $$ I_{1} = 0.482\,\mathrm{A} $$ The current in the primary winding is \(0.482\,\mathrm{A}\). In conclusion, the turns ratio of the transformer is 0.0085, and the currents in the primary and secondary windings are \(0.482\,\mathrm{A}\) and \(0.0041\,\mathrm{A}\), respectively.