
The primary winding of a transformer has $500\,$ turns where as its secondary has $5000$ turns. The primary is connected to an $a.c.$ supply of $20\,V$ , $50\,Hz$ . The secondary will have an output of
(A) $2\,V\,,5\,Hz$
(B) $20\,V\,,5\,Hz$
(C) $200\,V\,,50\,Hz$
(D) $20\,V\,,500\,Hz$
Answer
484.8k+ views
Hint:Write the relation between the number of turns in the windings and the output voltage and substitute the known values in it. The simplification of the above term, gives the answer for the output voltage and the frequency of the secondary winding.
Useful formula:
The relation between the number of turns and the voltage of the transformer is given by
$\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_s}}}{{{V_P}}}$
Where ${N_s}$ is the number of turns of the secondary winding, ${N_P}$ is the number of turns of the primary winding, ${V_s}$ is the voltage of the primary winding and ${V_P}$ is the voltage of the primary winding.
Complete step by step solution:
It is given that the
Number of turns in the primary winding of the transformer, ${N_P} = 500$
Number of turns in the secondary winding of the transformer, ${N_s} = 5000\,N$
The voltage of the primary coil, ${V_p} = 20\,V$
Using the relation of the transformer,
$\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_s}}}{{{V_P}}}$
By rearranging the above equation and substituting the known values in the above step, we get
\[\dfrac{{20}}{{{V_s}}} = \dfrac{{500}}{{5000}}\]
By grouping the similar terms together,
${V_s} = \dfrac{{20 \times 5000}}{{500}}$
By performing the basic arithmetic operation in the above step, we get
${V_s} = \dfrac{{100000}}{{500}}$
By simplifying the right hand side of the equation, we get
${V_s} = 200\,V$
The frequency of the windings are the same, and hence the secondary winding has the voltage output of $200\,V$ and the frequency of $50\,Hz$ .
Thus the option (C) is correct.
Note:The frequency is not influenced by the number of turns and it is the same for the primary and the secondary windings. The primary windings of the transformer collect the power from the battery and the secondary windings discharge the output voltage.
Useful formula:
The relation between the number of turns and the voltage of the transformer is given by
$\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_s}}}{{{V_P}}}$
Where ${N_s}$ is the number of turns of the secondary winding, ${N_P}$ is the number of turns of the primary winding, ${V_s}$ is the voltage of the primary winding and ${V_P}$ is the voltage of the primary winding.
Complete step by step solution:
It is given that the
Number of turns in the primary winding of the transformer, ${N_P} = 500$
Number of turns in the secondary winding of the transformer, ${N_s} = 5000\,N$
The voltage of the primary coil, ${V_p} = 20\,V$
Using the relation of the transformer,
$\dfrac{{{N_s}}}{{{N_P}}} = \dfrac{{{V_s}}}{{{V_P}}}$
By rearranging the above equation and substituting the known values in the above step, we get
\[\dfrac{{20}}{{{V_s}}} = \dfrac{{500}}{{5000}}\]
By grouping the similar terms together,
${V_s} = \dfrac{{20 \times 5000}}{{500}}$
By performing the basic arithmetic operation in the above step, we get
${V_s} = \dfrac{{100000}}{{500}}$
By simplifying the right hand side of the equation, we get
${V_s} = 200\,V$
The frequency of the windings are the same, and hence the secondary winding has the voltage output of $200\,V$ and the frequency of $50\,Hz$ .
Thus the option (C) is correct.
Note:The frequency is not influenced by the number of turns and it is the same for the primary and the secondary windings. The primary windings of the transformer collect the power from the battery and the secondary windings discharge the output voltage.
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