Question

A transformer having efficiency of 90% is working on $200\, V$ and $3 \,kW$ power supply. If the current in the secondary coil is $6\, A$, the voltage across the secondary coil and the current in the primary coil respectively are

• A. 300 V, 15 A
• B. 450 V, 15 A
• C. 450 V, 13.5 A
• D. 600 V, 15 A

Answer 1

The Correct Option is B

Here,
Efficiency of the transformer, $\eta=90\%$
Input power,$P_{in}=3 \, kW=3\times10^3 \, W=3000 \, W$
Voltage across the primary coil,$V_p=200 \, V$
Current in the secondary coil,$I_s=6 \, A$
As $P_{in}=I_pV_p$
$\therefore$ Current in the primary coil.
$I_p=\frac{P_{in}}{V_p}=\frac{3000\, W}{200 \, V}=15 \, A$
Efficiency of the transformer,
$\eta=\frac{P_{out}}{P_{in}}=\frac{V_sI_s}{V_pI_p}$
$\therefore \frac{90}{100}=\frac{6V_s}{3000} \, or \, V_s=\frac{90\times3000}{100\times6}=450\, V$

Answer 2

A transformer is a device that can increase or either decrease the voltage of the circuit as per the requirement. It consists of a primary coil and secondary coil. The primary coil acts as an input and the secondary coil acts as an output.
Both the coils have some power associated with them. The power of the input or the primary coil is passed to the secondary coil. In spite of that, there is some leakage of this power and so the output power (P_s) is less than the input power (P_p).
Hence, we define the efficiency of the transformer which is: η=\(\frac{P_{s}}{P_{p}}\)…. (i).
So, as It is given that the efficiency of the transformer is 90%, that means that the ratio is equal to 0.9. And the input power is given to be 3kW. This means P_p = 3kW = 3000W
By substituting the values of η and the input power in equation (i) we get,
⇒ 0.9 = \(\frac{P_{3}}{3000}\)
⇒ P_s = 0.9 × 3000 = 2700W
Hence, the power of the secondary coil = 2700W.
 So, let the voltage of the secondary coil and current flowing in the secondary coil be,
V_s and i_s respectively.
Hence, Power is equal to the product of the voltage and the current in the coil.
This gives us, P_s = V_si_s
It is given that, is = 6A and we found that P_s = 2700W.
So now this gives that, 2700 = 6V_s
Vs = \(\frac{2700}{6}\) = 450V
Thus, the voltage across the secondary coil is 450V.
It is also given that the voltage across the primary coil is 200V and the power in this coil is 3000W.
So, Let the voltage across the primary coil be V_p and the current in this coil be i_p.
Therefore now, P_p = V_pi_p.
So, by substituting the values of V_p and P_p
⇒ 3000 = 200i_p
⇒ i_p = 3000/200 = 15A
Hence, this means that the current in the primary coil is 15A.
Therefore, the correct option is ‘B’.  

Answer 3

P_in = 3x10³
V_pI_p = 3x10³
⇒ I_p = \(\frac{3\times 10^{3}}{200}\)
⇒ I_p = 15A
So now, 
η = \(\frac{P_{out}}{P_{in}}\)
⇒ \(\frac{V_{s}I_{s}}{3\times 10^{3}}\) = 90/100
⇒ V_s = 90 x 3 x \(\frac{10}{6}\)
⇒ V_s = 450V
Therefore the correct option is ‘B’ i.e 450V, 15A