Question

In a step-up transformer the voltage in the primary is 220Vand the current is 5A. The secondary voltage is found to be 22000V. The current in the secondary (neglect losses)is
A. 5A
B. 50A
C. 500A
D. 0.005A

Answer 1

Hint – If we are neglecting loses here so efficiency will be 100% So power of primary will be same for both the coil primary and secondary coil and we will use the formula of power in term of current and voltage ie; for both the coil primary and secondary.

Complete step by step solution
We have given voltage in primary coil is $220V$ and current is $5A$
Voltage in secondary coil is $22000V$
As we know, a transformer is a device which helps in the transformation of electric power in one circuit to electric power of the same frequency in another circuit. The voltage can be raised or lowered in a circuit.
So for step up transformer we can write
${N_s} > {N_p}$
Where Ns is no of turns in primary coils and $N_p$ is no of coils in secondary coil
If we neglected losses then efficiency will be 100% then power of primary and secondary will be same i.e.; ${P_p} = {P_s}$
We know that power is product of potential and current ie; P=VI
So in equation (4)
${V_p}{I_p} = {V_s}{I_s}$
Now putting the value of ${V_p}$ ,${I_p}$, ${V_s}$
$220 \times 5 = 22000{I_s}$
Now we need to isolate ${I_s}$
${I_s} = \dfrac{{220 \times 5}}{{22000}}$
=0.05A

Therefore option D is correct

Note:- for transformer ratio we have induced e.m.f $\varepsilon = \dfrac{{d\phi }}{{dt}}$
So for primary coil e.m.f ${\varepsilon _p} = {N_p}\dfrac{{d\phi }}{{dt}}$ And for secondary coil e.m.f ${\varepsilon _s} = {N_s}\dfrac{{d\phi }}{{dt}}$
Here flux are same so ratio of e.m.f will be
$\dfrac{{{N_s}}}{{{N_p}}} = \dfrac{{{\varepsilon _s}}}{{{\varepsilon _p}}}$
Substituting value of potential in equation
$\dfrac{{{N_s}}}{{{N_p}}} = 22000/220$=100