Question:
The coefficient of mutual induction between the primary and secondary coil of a transformer is $0.4\text{ H}$. When the current in the primary coil changes at the rate of $10\text{ As}^{-1}$, then the induced emf in the secondary will be:
The coefficient of mutual induction between the primary and secondary coil of a transformer is $0.4\text{ H}$. When the current in the primary coil changes at the rate of $10\text{ As}^{-1}$, then the induced emf in the secondary will be:
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Mutual induction relates the voltage in one coil to the rate of change of current in another.
The formula is a direct multiplication: $V = M \times (\text{rate of change of current})$.
The formula is a direct multiplication: $V = M \times (\text{rate of change of current})$.
Updated On: Jul 22, 2026
- $1.0\text{ V}$
- $4.0\text{ V}$
- $2.2\text{ V}$
- $3.1\text{ V}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the electromotive force (emf) induced in the secondary coil of a transformer when the current in the primary coil changes at a given rate.
Step 2: Key Formula and Approach:
By Faraday's Law of Electromagnetic Induction, the induced emf ($e_s$) in the secondary coil due to a changing current in the primary coil is:
\[ e_s = -M \frac{dI_p}{dt} \] where $M$ is the coefficient of mutual induction, and $\frac{dI_p}{dt}$ is the rate of change of current in the primary coil.
We will find the magnitude of the induced emf.
Step 3: Detailed Explanation:
• Identify given values:
Mutual inductance $M = 0.4\text{ H}$
Rate of change of primary current $\frac{dI_p}{dt} = 10\text{ As}^{-1}$
• Calculate the magnitude of induced emf ($|e_s|$):
\[ |e_s| = M \left| \frac{dI_p}{dt} \right| \] \[ |e_s| = 0.4\text{ H} \times 10\text{ As}^{-1} \] \[ |e_s| = 4\text{ V} \]
Step 4: Final Answer:
The induced emf in the secondary coil is $4.0\text{ V}$, which corresponds to Option (B).
The question asks for the electromotive force (emf) induced in the secondary coil of a transformer when the current in the primary coil changes at a given rate.
Step 2: Key Formula and Approach:
By Faraday's Law of Electromagnetic Induction, the induced emf ($e_s$) in the secondary coil due to a changing current in the primary coil is:
\[ e_s = -M \frac{dI_p}{dt} \] where $M$ is the coefficient of mutual induction, and $\frac{dI_p}{dt}$ is the rate of change of current in the primary coil.
We will find the magnitude of the induced emf.
Step 3: Detailed Explanation:
• Identify given values:
Mutual inductance $M = 0.4\text{ H}$
Rate of change of primary current $\frac{dI_p}{dt} = 10\text{ As}^{-1}$
• Calculate the magnitude of induced emf ($|e_s|$):
\[ |e_s| = M \left| \frac{dI_p}{dt} \right| \] \[ |e_s| = 0.4\text{ H} \times 10\text{ As}^{-1} \] \[ |e_s| = 4\text{ V} \]
Step 4: Final Answer:
The induced emf in the secondary coil is $4.0\text{ V}$, which corresponds to Option (B).
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