Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACTransformerEasy2 minPYQ_2020

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An ideal transformer converts $220 V$ AC to $3.3 kV$ AC. It transmits the power of $4.4 kW$ . If the primary coil has $600$ turns, then alternating current in the secondary coil is

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Solution:

$V_{i n} = 220 V, V_{o u t} = 3.3 \times 10^{3} V$

$P o w e r = 4.4 k W$

$N_{p} = 600$

$P = V_{i n} \times I_{i n}$ .

$I_{i n} = \frac{4.4 \times 1000}{220} = \frac{44 \times 10}{22} = 20 A$

$\frac{e_{s}}{e_{p}} = \frac{I_{p}}{I_{s}}$

$I_{s} = I_{p} = \frac{e_{p}}{e_{s}} \times \frac{20 \times 220}{3.3 \times 1000} = \frac{44}{33} = \frac{4}{3} A$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Transformer

⏱ 2mℹ️ Source: PYQ_2020

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