Electrical Engineering - Transformers - Discussion
Discussion Forum : Transformers - General Questions (Q.No. 4)
4.
If a transformer has 50 turns in the primary winding and 10 turns in the secondary winding, what is the reflective resistance if the secondary load resistance is 250 
?
?250
25
6,250
62,500
Discussion:
46 comments Page 2 of 5.
Faisal said:
8 years ago
The Formula is this, R1=(N1/N2)^2 RL.
and R1 is given, N1 is given, N2 is given,
So, 250=(50/10)^2 RL.
250 = 25 RL.
Therefore, RL = 250 * 25 = 6250.
and R1 is given, N1 is given, N2 is given,
So, 250=(50/10)^2 RL.
250 = 25 RL.
Therefore, RL = 250 * 25 = 6250.
(2)
ROCKY said:
8 years ago
Transformation ratio is N2/N1. Am I right?
(1)
Arjun said:
9 years ago
Reflective resistance = {(1/k^2)*Rload}.
Sandeep Kumar said:
9 years ago
N2/N1= Sq.rt(R2/R1)
10/50 = Sq.Rt(250/R1)
(1/5)^2 = 250/R1.
R1=250*25 = 6250.
10/50 = Sq.Rt(250/R1)
(1/5)^2 = 250/R1.
R1=250*25 = 6250.
(1)
Vyankatesh badgujar said:
9 years ago
N1= 50 , N2 = 10.
Turn ratio K = N2/N1 = 10/50,
K = 1/5 = 0.2,
Now, R2/R1 = K^2.
R1 = R2/K^2,
= 250/0.2^2,
R1 = 6250 ohms.
Turn ratio K = N2/N1 = 10/50,
K = 1/5 = 0.2,
Now, R2/R1 = K^2.
R1 = R2/K^2,
= 250/0.2^2,
R1 = 6250 ohms.
(1)
Santosh murkut said:
9 years ago
The transformation ratio is said that N2/N1 = V2/V1 = I1/I2 = K.
Hence,
1) A resistance R1 in primary become K^2 * R1 when transferred to the secondary.
2) A resistance R2 in secondary become R2 / K^2 when transferred to the primary.
Therefore,
R1 = 250 / (10/50)^2 = 6250 ohm.
Hence,
1) A resistance R1 in primary become K^2 * R1 when transferred to the secondary.
2) A resistance R2 in secondary become R2 / K^2 when transferred to the primary.
Therefore,
R1 = 250 / (10/50)^2 = 6250 ohm.
Rudra said:
1 decade ago
What will be the Phasor diagram for it?
Bhawana Singh said:
1 decade ago
Turn ratio will proportional to square of load then square of resistance in secondary upon primary 2n square.
Sandip chhatrola said:
1 decade ago
K^2 = R2/R1.
1/25 = 250/R1.
R1 = 6250 ohm.
1/25 = 250/R1.
R1 = 6250 ohm.
Akash.bhure said:
1 decade ago
Solution:
No.of turns in the primary(N1) = 50.
No.of turns in the secondary(N2) = 10.
Turns ratio(K) = N2/N1 = 10/50 = 0.2.
Secondary load resistance(R2) = 250 Ohm.
Reflective resistance(R1) = R2/K^2.
(R1) = 250/(0.2)^2.
(R1) = 250/0.04 = 6250 ohm's.
No.of turns in the primary(N1) = 50.
No.of turns in the secondary(N2) = 10.
Turns ratio(K) = N2/N1 = 10/50 = 0.2.
Secondary load resistance(R2) = 250 Ohm.
Reflective resistance(R1) = R2/K^2.
(R1) = 250/(0.2)^2.
(R1) = 250/0.04 = 6250 ohm's.
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