Question

In a transformer without any loss in any power, there are $5000$ turns in the primary and $250$ turns in the secondary. The primary voltage is $120V$ and the primary current is $0.1A$. Find the voltage and current in the secondary?

Answer 1

Hint: Since we can see that there is no power loss. So, from this, we can say that the total induced voltage in each winding will be directly proportional to the number of turns in that particular winding. Therefore, the number of turns and the current in the primary and secondary windings are related by an inverse proportion. So by using the formula which is $\dfrac{{{N_p}}}{{{N_S}}} = \dfrac{{{V_p}}}{{{V_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$ and substituting the values, we will get the voltages and current in the secondary coil.
Formula used:
Primary and secondary coil relationship given by,
$\dfrac{{{N_p}}}{{{N_S}}} = \dfrac{{{V_p}}}{{{V_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$
Here,
${N_P}$, is the number of turns in the primary coil
${N_S}$, is the number of turns in the secondary coil
${V_P}$, is the primary voltage
${V_S}$, is the secondary voltage
${I_P}$, is the primary current
${I_S}$, is the secondary current

Complete step by step answer: First of all we will see the values known to us and what we have to find. So in the question, the values given to us are:
${N_P} = 5000$
${N_S} = 250$
${V_P} = 120V$
${I_P} = 0.1A$
So, we have to find ${V_S}$ and ${I_S}$. For this, we will use the relation mentioned above, which is given as
$\dfrac{{{N_p}}}{{{N_S}}} = \dfrac{{{V_p}}}{{{V_S}}} = \dfrac{{{I_S}}}{{{I_P}}}$
And the above relation can also be written as
$ \Rightarrow \dfrac{{{N_p}}}{{{N_S}}} = \dfrac{{{V_p}}}{{{V_S}}}$
Now on substituting the values in the above relation, we get
\[ \Rightarrow \dfrac{{5000}}{{250}} = \dfrac{{120}}{{{V_S}}}\]
On solving the numerator and denominator part and reducing it in the simplest form, we get
\[ \Rightarrow 20 = \dfrac{{120}}{{{V_S}}}\]
Now taking the constant term one side and solving the equation, we get
$ \Rightarrow {V_s} = 6V$
Therefore, the secondary voltage of the coil will be $6V$ .
Now we have to find current in the secondary coil, so for this, we will write the relation as
\[ \Rightarrow \dfrac{{{N_p}}}{{{N_S}}} = \dfrac{{{I_S}}}{{{I_P}}}\]
Now on substituting the values in the above relation, we get
\[ \Rightarrow \dfrac{{5000}}{{250}} = \dfrac{{{I_S}}}{{0.1}}\]
Now by reducing the equation in the simplest form, we get the equation as
\[ \Rightarrow 20 = \dfrac{{{I_S}}}{{0.1}}\]
On taking the cross-multiplication, we get
\[ \Rightarrow {I_S} = 2A\]

Therefore, the secondary current of the coil will be $2A$.

Note: So, to verify this solution whether it is right or wrong we should remember this. As in a single-phase voltage transformer, the primary coil is usually the side with the higher voltage as compared to the secondary coil. And also we should always mention the unit in the answer.