
It is required to construct a transformer which gives 12 V from a 240 V AC supply
The number of turns in the primary is 4800.
a) Calculate the number of turns in the secondary.
b) State weather, coil in the secondary has to be thick or thin. Justify your answer.
Answer
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Hint: The transformer given in the question is a step up transformer because voltage increases and current decreases in order to maintain the constant power supply of the transformer.
So, The formulae used in the calculation,
$\dfrac{{{V_S}}}{{{V_P}}} = \dfrac{{{N_S}}}{{{N_P}}}$
$P = V \times I$
Complete step by step solution:
(a) Given : ${V_P}$ = 240 V, ${V_S}$ = 12 V and ${N_P}$ = 4800
Let the ${V_S}$ and ${V_P}$ are the A.C. voltage of secondary and primary coil respectively & ${N_P}$and ${N_S}$ are the number of turns in the primary and secondary coil respectively.
According to formula
$\dfrac{{{V_S}}}{{{V_P}}} = \dfrac{{{N_S}}}{{{N_P}}}$
$\dfrac{{12}}{{240}} = \dfrac{{{N_S}}}{{4800}} \Leftrightarrow {N_S} = \dfrac{{12}}{{240}} \times 4800$
On solving we get, ${N_S} = 240$
Hence the number of turns in the secondary coil will be 240.
(b) We know that the voltage increases from low to high therefore it is a step up transformer. As the transformer is step up, the secondary voltage is more than the primary voltage but the power will remain constant so in order to do that the current in the secondary coil will be less than the primary coil in accordance with the relation,
$P = {V_s} \times {I_s}$
If ${V_S}$ increases then ${I_{_S}}$will have to decrease for the power to remain constant and if current decreases then for the flow of current thin wire is sufficient so that heat dissipation would be minimal. Hence we need to use thin wire for the step up transformer.
Note: The number of turns in a coil is directly proportional to voltage but inversely proportional to current in the coil by the conservation of power.
For step up transformer thin wire is required whereas for step down transformer thick wire is required for power supply in transformers.
So, The formulae used in the calculation,
$\dfrac{{{V_S}}}{{{V_P}}} = \dfrac{{{N_S}}}{{{N_P}}}$
$P = V \times I$
Complete step by step solution:
(a) Given : ${V_P}$ = 240 V, ${V_S}$ = 12 V and ${N_P}$ = 4800
Let the ${V_S}$ and ${V_P}$ are the A.C. voltage of secondary and primary coil respectively & ${N_P}$and ${N_S}$ are the number of turns in the primary and secondary coil respectively.
According to formula
$\dfrac{{{V_S}}}{{{V_P}}} = \dfrac{{{N_S}}}{{{N_P}}}$
$\dfrac{{12}}{{240}} = \dfrac{{{N_S}}}{{4800}} \Leftrightarrow {N_S} = \dfrac{{12}}{{240}} \times 4800$
On solving we get, ${N_S} = 240$
Hence the number of turns in the secondary coil will be 240.
(b) We know that the voltage increases from low to high therefore it is a step up transformer. As the transformer is step up, the secondary voltage is more than the primary voltage but the power will remain constant so in order to do that the current in the secondary coil will be less than the primary coil in accordance with the relation,
$P = {V_s} \times {I_s}$
If ${V_S}$ increases then ${I_{_S}}$will have to decrease for the power to remain constant and if current decreases then for the flow of current thin wire is sufficient so that heat dissipation would be minimal. Hence we need to use thin wire for the step up transformer.
Note: The number of turns in a coil is directly proportional to voltage but inversely proportional to current in the coil by the conservation of power.
For step up transformer thin wire is required whereas for step down transformer thick wire is required for power supply in transformers.
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