JEE Challenger
More from Alternating Current
PhyHardAdv 2013 • Paper 2

Transformer Turns Ratio Calculation for Power Transmission

Comprehension Passage

A thermal power plant produces electric power of 600 kW600\text{ kW} at 4000 V4000\text{ V}, which needs to be delivered to a facility 20 km20\text{ km} away for consumer usage. Power can be transported either directly using a high current carrying capacity cable, or through a system involving step-up and step-down transformers at opposite ends. Direct transmission leads to significant energy loss through dissipation, whereas the transformer method reduces dissipation drastically. In this process, a step-up transformer at the power station reduces the transmission current. At the consumer end, a step-down transformer reduces the voltage to the required lower value. Assume the transmission cable is strictly resistive, all transformers are ideal with a power factor of unity, and all specified values of voltage and current are root-mean-square (rms) values.

In the transformer-assisted transmission scheme, suppose the ratio of primary to secondary turns of the step-up transformer is 1:101 : 10. If the power supplied to the end consumers is at a voltage of 200 V200\text{ V}, find the ratio of primary to secondary turns of the step-down transformer.

Options

A

200:1200 : 1

Correct
B

150:1150 : 1

C

100:1100 : 1

D

50:150 : 1

Topics & Concepts

Alternating CurrentTransformers

Step-by-Step Solution

To find the ratio of primary to secondary turns of the step-down transformer, we follow a step-by-step calculation:

Step 1: Step-up Transformer Analysis

The power plant generates electric power at a voltage of Vp1=4000 VV_{p1} = 4000\text{ V}. The step-up transformer has a primary to secondary turns ratio given by: Np1Ns1=110\frac{N_{p1}}{N_{s1}} = \frac{1}{10}

Since the transformer is ideal, the output voltage across the secondary winding Vs1V_{s1} is given by: Vs1=Vp1×(Ns1Np1)=4000 V×10=40000 VV_{s1} = V_{p1} \times \left(\frac{N_{s1}}{N_{p1}}\right) = 4000\text{ V} \times 10 = 40000\text{ V}

Step 2: Transmission Line Voltage Drop

The total power transmitted is P=600 kW=600×103 WP = 600\text{ kW} = 600 \times 10^3\text{ W}.

  1. Transmission Current (II): I=PVs1=600×103 W40000 V=15 AI = \frac{P}{V_{s1}} = \frac{600 \times 10^3\text{ W}}{40000\text{ V}} = 15\text{ A}

  2. Resistance of Transmission Line (RR): The distance is 20 km20\text{ km}, and using the resistance rate 0.4 Ω/km0.4\text{ }\Omega/\text{km} (from Question 15): R=20 km×0.4 Ω/km=8 ΩR = 20\text{ km} \times 0.4\text{ }\Omega/\text{km} = 8\text{ }\Omega

  3. Voltage Drop (ΔV\Delta V): ΔV=I×R=15 A×8 Ω=120 V\Delta V = I \times R = 15\text{ A} \times 8\text{ }\Omega = 120\text{ V}

  4. Voltage at Primary of Step-down Transformer (Vp2V_{p2}): Vp2=Vs1−ΔV=40000 V−120 V=39880 VV_{p2} = V_{s1} - \Delta V = 40000\text{ V} - 120\text{ V} = 39880\text{ V}

Step 3: Step-down Transformer Turns Ratio

The consumer end requires a output voltage of Vs2=200 VV_{s2} = 200\text{ V}.

The ratio of primary to secondary turns of the step-down transformer is: Np2Ns2=Vp2Vs2=39880 V200 V=199.4\frac{N_{p2}}{N_{s2}} = \frac{V_{p2}}{V_{s2}} = \frac{39880\text{ V}}{200\text{ V}} = 199.4

Rounding to the nearest whole integer (or neglecting the minor transmission line drop 120 V≪40000 V120\text{ V} \ll 40000\text{ V}): Np2Ns2≈40000 V200 V=200\frac{N_{p2}}{N_{s2}} \approx \frac{40000\text{ V}}{200\text{ V}} = 200

Thus, the turns ratio is 200:1200 : 1.

Correct Answer: A

PreviousNext