Question

The efficiency of a 10kVa , 2000/400 V single phase transformer at unity pf is 97% at rated load and also half rated load . determine the transformer core losses and ohmic loss.

Answer 1

Answer:

Answer:

Perfect questions for guys interested in maths..

2. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,    

2 + 2(22)

= 2 + 2(2 × 2)

= 2 + 2(4)

= 2 + 8

= 10

Answer: (a

Explanation:

Answer 2

Answer:

Transformer core loss is $103.1$ $W$. Ohmic loss is $206.185$ $W$.

Explanation:

• The work of the transformer is to transfer electric signals from one AC circuit to the other AC circuit by either decreasing voltage or increasing voltage.
• Transformers can be classified as step-up transformers or step-down transformers.

The formula to calculate efficiency is given as:

$Efficiency=\frac{X_{out} }{X_{in} }$

                   $= 1-\frac{X_{L} }{X_{in} }$

where, $X_{out}$ is the output.

            $X_{in}$ is the input.

            $X_{L}$ is the loss.

Step 1:

At full load:

$0.97=1-\frac{Pc +Pcw}{X_{out} + Pc +Pcw }$

where, $Pc$ is the core loss.

            $Pcw$ is the ohmic copper loss.

$0.97=1-\frac{Pc +Pcw}{10*10^{3}*1 + Pc +Pcw }$

$\frac{Pc +Pcw}{10000 + Pc +Pcw }=0.03$

$Pc + Pcw=309.278$ $W$

Step 2:

At half load:

$0.97=1-\frac{Pc +\frac{1}{4} Pcw}{10 *10^{3}*\frac{1}{2} + Pc +\frac{1}{4} Pcw }$

$Pc +\frac{1}{4}Pcw=150 + 0.03 (Pc +\frac{1}{4}Pcw)$

$Pc +\frac{1}{4}Pcw=\frac{150}{0.97}$

$Pc +\frac{1}{4}Pcw=154.639$ $W$

Solving the two equations we get:

$Pc=103.1$ $W$

$Pcw=206.185$ $W$

So, transformer core loss is $103.1$ $W$. Ohmic loss is $206.185$ $W$.