
A transformer with primary to secondary turns ratio of $1:2$, is connected to an alternator of voltage $200\,V$. A current of $4\,A$ is flowing through the primary coil. Assuming that the transformer has no losses, the secondary voltage and current are respectively $:$
$
A.\,100\,V,\,8\,A \\
B.\,400\,V,8\,A \\
C.\,400\,V,2\,A \\
D.\,100\,V,\,2\,A \\
$
Answer
571.8k+ views
Hint: From the question, we know that the values of turns in the coil, primary voltage and current of the coils. Substituting the known values in the equation of turns ratio we get the value of the secondary voltage and current of the coil.
Formulae Used:
The expression for finding the turns in the coil is
$\dfrac{{{i_p}}}{{{i_s}}} = \,\dfrac{{{v_p}}}{{{v_s}}}$
Where ${i_p}$ be the number of turns in the primary coil, ${i_s}$be the number of turns in the secondary coil, ${v_p}$ be the potential of primary voltage and ${v_s}$ be the potential of secondary voltage.
Complete step-by-step solution:
From the question we know that,
Primary to secondary ratio is $1:2$
Current flowing through the primary coil \[\,({i_p})\, = \,4\,A\]
Potential of primary voltage $({V_p}) = \,200\,V$
$\dfrac{{{i_p}}}{{{i_s}}} = \,\dfrac{{{v_p}}}{{{v_s}}}........\left( 1 \right)$
Substitute the known values in the equation $\left( 1 \right)$
$\dfrac{1}{2} = \dfrac{4}{{{i_s}}}$
Simplify the above equation we get,
${i_s} = 2\,A$
We know that,
$Power\, = \,voltage\, \times \,current$
Equating the power equation we get,
${i_s} \times \,{v_s}\, = \,{i_p}\, \times \,{v_p}$
Substitute the known values in the above equation we get,
$2 \times \,{v_s} = 4 \times 200$
${v_s} = 400\,A$
Therefore, voltage and current of secondary voltage are $400\,A\,and\,2A$.
Hence from the above option, option $\left( C \right)$ is correct.
Note:- From the question, we know that the transformer has no losses so we no need to find the transformer loss. we know that the power is proportional to the voltage and current. In the question we use the equation called ratio of transformation. This is also known as turns ratio.
Formulae Used:
The expression for finding the turns in the coil is
$\dfrac{{{i_p}}}{{{i_s}}} = \,\dfrac{{{v_p}}}{{{v_s}}}$
Where ${i_p}$ be the number of turns in the primary coil, ${i_s}$be the number of turns in the secondary coil, ${v_p}$ be the potential of primary voltage and ${v_s}$ be the potential of secondary voltage.
Complete step-by-step solution:
From the question we know that,
Primary to secondary ratio is $1:2$
Current flowing through the primary coil \[\,({i_p})\, = \,4\,A\]
Potential of primary voltage $({V_p}) = \,200\,V$
$\dfrac{{{i_p}}}{{{i_s}}} = \,\dfrac{{{v_p}}}{{{v_s}}}........\left( 1 \right)$
Substitute the known values in the equation $\left( 1 \right)$
$\dfrac{1}{2} = \dfrac{4}{{{i_s}}}$
Simplify the above equation we get,
${i_s} = 2\,A$
We know that,
$Power\, = \,voltage\, \times \,current$
Equating the power equation we get,
${i_s} \times \,{v_s}\, = \,{i_p}\, \times \,{v_p}$
Substitute the known values in the above equation we get,
$2 \times \,{v_s} = 4 \times 200$
${v_s} = 400\,A$
Therefore, voltage and current of secondary voltage are $400\,A\,and\,2A$.
Hence from the above option, option $\left( C \right)$ is correct.
Note:- From the question, we know that the transformer has no losses so we no need to find the transformer loss. we know that the power is proportional to the voltage and current. In the question we use the equation called ratio of transformation. This is also known as turns ratio.
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