Question

When a current of $5\text{ A}$ passes through the primary coil of a transformer of $200$ number turns, the magnetic flux linked with the secondary coil having $400$ turns is $600 \times 10^{-6}\text{ T m}^2$. Find the induced emf in the secondary coil, when the current in the primary coil increases at a rate of $2\text{ A s}^{-1}$.

• A. $0.92 \times 10^{-4}\text{ V}$
• B. $1.92 \times 10^{-4}\text{ V}$
• C. $0.92 \times 10^{-2}\text{ V}$
• D. $1.92 \times 10^{-2}\text{ V}$

Hint:

Notice that the total number of primary turns ($N_p = 200$) is extra information provided to confuse you. The relationship for mutual inductance strictly depends on the flux lines cutting through the secondary turns due to the primary current source. Always keep an eye out for redundant data!

Answer 1

The Correct Option is D

Concept: The total magnetic flux linkage through a secondary inductor coil ($\phi_s$) due to a current passing in the primary loop ($I_p$) is related by the mutual inductance factor $M$: \[ N_s \phi_s = M I_p \] According to Faraday's law of electromagnetic induction, the magnitude of the electromotive force (emf) induced in the secondary inductor as the primary current changes over time is: \[ e_s = M \frac{dI_p}{dt} \]

Step 1: Determine the mutual inductance ($M$).
We are given the following values:
• Primary current, $I_p = 2.5\text{ A}$
• Secondary turns, $N_s = 400$
• Flux through a single secondary turn, $\phi_s = 600 \times 10^{-6}\text{ T m}^2$
• Rate of change of primary current, $\frac{dI_p}{dt} = 0.2\text{ A s}^{-1}$ Using the flux linkage relation to isolate $M$: \[ M = \frac{N_s \phi_s}{I_p} \] \[ M = \frac{400 \times 600 \times 10^{-6}}{2.5} = \frac{240000 \times 10^{-6}}{2.5} = 96000 \times 10^{-6} = 0.096\text{ H} \]

Step 2: Calculate the induced secondary emf ($e_s$).
Using Faraday's statement: \[ e_s = M \frac{dI_p}{dt} \] \[ e_s = 0.096 \times 0.2 = 0.0192\text{ V} = 1.92 \times 10^{-2}\text{ V} \]