Find the current through the primary coil \((P)\) of the transformer shown below.
- \(0.08 \mathrm{~A}\)
- \(0.04 \mathrm{~A}\)
- \(0.02 \mathrm{~A}\)
- \(0.01 \mathrm{~A}\)
Solution
For the given transformer, \(\begin{aligned} & V_p=230 \mathrm{~V} \\ & V_S=23 \mathrm{~V} \\ & R_S=115 \Omega \end{aligned}\) Currrent in secondary coil, \(I_S=\frac{V_S}{R_S}=\frac{23}{115}=0.2 \mathrm{~A}\) We know that, in a transformer \(\begin{aligned} \frac{V_S}{V_P} & =\frac{I_P}{I_S} \\ \Rightarrow \quad & I_P=\frac{V_S I_S}{V_P}=\frac{23 \times 0.2}{230}=0.02 \mathrm{~A} \end{aligned}\)
Asked in: AP EAMCET 2020 (21 Sep Shift 1)