Voltage in the secondary coil of a transforme | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between the voltage in the primary coil $(V_P)$ and the secondary coil $(V_S)$ of an ideal transformer is given by the transforme

Vedclass · Accepted Answer

Frequency of the source

Vedclass · Answer

(C) The relationship between the voltage in the primary coil $(V_P)$ and the secondary coil $(V_S)$ of an ideal transformer is given by the transformer equation:$\frac{V_S}{V_P} = \frac{N_S}{N_P}$Where $N_S$ is the number of turns in the secondary coil and $N_P$ is the number of turns in the primary coil.From this equation,it is clear that $V_S = V_P 	imes (N_S / N_P)$. Therefore,the secondary voltage depends on the primary voltage and the turns ratio $(N_S / N_P)$.The frequency of the source does not appear in this ideal transformer equation,meaning the secondary voltage is independent of the source frequency.

Voltage in the secondary coil of a transformer does not depend upon.

Similar Questions

In a step-up transformer,the transformation ratio is $4:1$. If a Leclanche cell having an $EMF$ of $1.5 \ V$ is connected to the primary coil of the transformer,then the voltage obtained across its secondary coil is . . . . . . .

Which device is used to transmit voltage over large distances and why?

$A$ $60\, W$ load is connected to the secondary of a transformer whose primary draws line voltage. If a current of $0.54\, A$ flows in the load,what is the current in the primary coil? Comment on the type of transformer being used.

Using the equation of power for an ideal transformer,prove $\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$.

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 the alternating current in the secondary coil is

Vedclass Test Series

Exam Paper Generator

Online Exam Module