Understanding Potential Transformer (PT) and Current Transformer (CT) Usage

In electrical power systems, Potential Transformers (PTs) and Current Transformers (CTs) are used to reduce high voltages and currents, respectively, to safe levels for measuring instruments like voltmeters and ammeters. This question involves using the known ratios of these transformers and the readings from the instruments to determine the actual voltage and current in the transmission line.

Calculating Transmission Line Voltage

A Potential Transformer (PT) steps down the high transmission line voltage to a lower, measurable voltage. The ratio of the primary voltage (line voltage) to the secondary voltage (voltmeter reading) is constant.

The given PT ratio is $\frac{12000 \text{ V}}{112 \text{ V}}$.

The voltmeter reading ($V_{meter}$) is given as $120 \text{ V}$. This reading is taken across the secondary winding of the PT.

We can set up a proportion to find the actual line voltage ($V_{line}$):

$$ \frac{V_{line}}{V_{meter}} = \frac{\text{Primary Voltage Rating of PT}}{\text{Secondary Voltage Rating of PT}} $$

Substituting the known values:

$$ \frac{V_{line}}{120 \text{ V}} = \frac{12000 \text{ V}}{112 \text{ V}} $$

Now, we solve for $V_{line}$:

$$ V_{line} = 120 \text{ V} \times \frac{12000}{112} $$

$$ V_{line} = 120 \times 107.1428... $$

$$ V_{line} \approx 12857.14 \text{ V} $$

Calculating Transmission Line Current

Similarly, a Current Transformer (CT) steps down the high transmission line current to a lower, measurable current. The ratio of the primary current (line current) to the secondary current (ammeter reading) is constant.

The given CT ratio is $\frac{80 \text{ A}}{6 \text{ A}}$.

The ammeter reading ($I_{ammeter}$) is given as $5 \text{ A}$. This reading is taken from the secondary winding of the CT.

We can set up a proportion to find the actual line current ($I_{line}$):

$$ \frac{I_{line}}{I_{ammeter}} = \frac{\text{Primary Current Rating of CT}}{\text{Secondary Current Rating of CT}} $$

Substituting the known values:

$$ \frac{I_{line}}{5 \text{ A}} = \frac{80 \text{ A}}{6 \text{ A}} $$

Now, we solve for $I_{line}$:

$$ I_{line} = 5 \text{ A} \times \frac{80}{6} $$

$$ I_{line} = 5 \times 13.3333... $$

$$ I_{line} \approx 66.67 \text{ A} $$

Conclusion

Based on the calculations using the Potential Transformer and Current Transformer ratios, the voltage in the transmission line is approximately $12857$ V and the current is approximately $66.67$ A.