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A $60 \mathrm{~W}$ load is connected to the secondary of a transformer whose primary draws line voltage. If a current of $0.54 \mathrm{~A}$ flows in the load, what is the current in the primary coil? Comment on the type of tansformer being used.
Solution:
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Verified Answer
As given that,
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W} \\
&\mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A}
\end{aligned}
$$
Primary current $\mathrm{I}_{\mathrm{p}}=$ ?
Taking line voltage as $220 \mathrm{~V}\left(\mathrm{~V}_{\mathrm{P}}=220 \mathrm{~V}\right)$
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} \\
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W}, \mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A} \\
&\mathrm{~V}_{\mathrm{S}}=\frac{60}{0.54}=110 \mathrm{Volt} .
\end{aligned}
$$
Voltage in the secondary $\left(\mathrm{V}_{\mathrm{S}}\right)$ is less than voltage in the primary $\left(\mathrm{V}_{\mathrm{p}}\right)$.
So, the transformer is step down transformer.
since, the transformation ratio factor is
$$
\begin{aligned}
&r=\frac{\text { Secondary voltage }}{\text { Primary voltage }} \\
&r=\frac{V_S}{V_P}
\end{aligned}
$$
$\therefore$ For transformer $\frac{V_S}{V_{\mathrm{p}}}=\frac{I_P}{I_s}$
Substituting the values,
$$
\begin{aligned}
\frac{110 \mathrm{~V}}{220 \mathrm{~V}} &=\frac{\mathrm{I}_{\mathrm{P}}}{0.54 \mathrm{~A}} \\
\Rightarrow \quad \quad \mathrm{I}_{\mathrm{P}} &=0.27 \text { Ampere }
\end{aligned}
$$
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W} \\
&\mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A}
\end{aligned}
$$
Primary current $\mathrm{I}_{\mathrm{p}}=$ ?
Taking line voltage as $220 \mathrm{~V}\left(\mathrm{~V}_{\mathrm{P}}=220 \mathrm{~V}\right)$
$$
\begin{aligned}
&\mathrm{P}_{\mathrm{S}}=\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}} \\
&\mathrm{P}_{\mathrm{S}}=60 \mathrm{~W}, \mathrm{I}_{\mathrm{S}}=0.54 \mathrm{~A} \\
&\mathrm{~V}_{\mathrm{S}}=\frac{60}{0.54}=110 \mathrm{Volt} .
\end{aligned}
$$
Voltage in the secondary $\left(\mathrm{V}_{\mathrm{S}}\right)$ is less than voltage in the primary $\left(\mathrm{V}_{\mathrm{p}}\right)$.
So, the transformer is step down transformer.
since, the transformation ratio factor is
$$
\begin{aligned}
&r=\frac{\text { Secondary voltage }}{\text { Primary voltage }} \\
&r=\frac{V_S}{V_P}
\end{aligned}
$$
$\therefore$ For transformer $\frac{V_S}{V_{\mathrm{p}}}=\frac{I_P}{I_s}$
Substituting the values,
$$
\begin{aligned}
\frac{110 \mathrm{~V}}{220 \mathrm{~V}} &=\frac{\mathrm{I}_{\mathrm{P}}}{0.54 \mathrm{~A}} \\
\Rightarrow \quad \quad \mathrm{I}_{\mathrm{P}} &=0.27 \text { Ampere }
\end{aligned}
$$
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