A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of 15 Ω the currents in the primary and secondary respectively, are

Question

A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of 15 Ω the currents in the primary and secondary respectively, are (A) 0.16 A , 3.2 × 10-3 A (B) 3.2 × 10-3 A , 0.16 A (C) 0.16 A , × 0.16 A (D) 3.2 × 10 .-3 A , 3.2 × 10-3 A

Tardigrade Admin · Accepted Answer

We have, (NS/NP)=(iP/IS) (10/500)=(iS/iS) ⇒ (iP/iS)=(1/50) ⇒ is=50 ip This condition is satisfied only when current in primary 3.2 × 10-3 A and in secondary 0.16 A.

Q. A transformer has $500$ primary turns and $10$ secondary turns. If the secondary has a resistive load of $15 Ω$ the currents in the primary and secondary respectively, are

$0.16 A, 3.2 \times 1 0^{- 3} A$

$3.2 \times 1 0^{- 3} A, 0.16 A$

$0.16 A, \times 0.16 A$

$3.2 \times 10 .^{- 3} A, 3.2 \times 1 0^{- 3} A$

We have, $\frac{N _{S}}{N _{P}} = \frac{i _{P}}{I _{S}} \frac{10}{500} = \frac{i _{S}}{i _{S}}$ $\Rightarrow \frac{i _{P}}{i _{S}} = \frac{1}{50}$
$\Rightarrow i_{s} = 50 i_{p}$
This condition is satisfied only when current in primary
$3.2 \times 1 0^{- 3} A$ and in secondary $0.16 A$ .

Q. A transformer has 500 primary turns and 10 secondary turns. If the secondary has a resistive load of 15Ω the currents in the primary and secondary respectively, are

0.16A,3.2×10−3A

3.2×10−3A,0.16A

0.16A,×0.16A

3.2×10.−3A,3.2×10−3A