In a step-up transformer the voltage in the primary is 220 V and die current is 5 A . The secondary voltage is found to be 22000 V . The current in the secondary (neglect losses) is

Question

In a step-up transformer the voltage in the primary is 220 V and die current is 5 A . The secondary voltage is found to be 22000 V . The current in the secondary (neglect losses) is (A) 5A (B) 50A (C) 500A (D) 0.05A

Tardigrade Admin · Accepted Answer

For a transformer, (Vs/Vp)=(ip/is) ⇒ is=ip× (VpV/s)=5× (220/22000) =0.05 A

Q. In a step-up transformer the voltage in the primary is $220 V$ and die current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

5A

50A

500A

0.05A

For a transformer, $\frac{V _{s}}{V _{p}} = \frac{i _{p}}{i _{s}}$
$\Rightarrow i_{s} = i_{p} \times \frac{V _{p}}{V}_{s} = 5 \times \frac{220}{22000}$
$= 0.05 A$

Q. In a step-up transformer the voltage in the primary is 220V and die current is 5A . The secondary voltage is found to be 22000V . The current in the secondary (neglect losses) is

5A

50A

500A

0.05A

For a transformer, Vp​Vs​​=is​ip​​ ⇒is​=ip​×VVp​​s​=5×22000220​ =0.05A

Q. In a step-up transformer the voltage in the primary is $220 V$ and die current is $5 A$ . The secondary voltage is found to be $22000 V$ . The current in the secondary (neglect losses) is

For a transformer, $\frac{V _{s}}{V _{p}} = \frac{i _{p}}{i _{s}}$
$\Rightarrow i_{s} = i_{p} \times \frac{V _{p}}{V}_{s} = 5 \times \frac{220}{22000}$
$= 0.05 A$