Transformers: Practice Problem & Solution
The primary and secondary coil of a transformer have 50 and 1500 turns respectively. If the magnetic flux $\phi$ linked with the primary coil is given by $\phi = \phi_0 - 4t$, where $\phi$ is in Webers, t is time in seconds and $\phi_0$ is a constant, the output voltage across the secondary coil is: (2007)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Transformers. Understanding the underlying formula is key to arriving at the correct answer below:
Voltage across primary is $V_p = \left| \frac{d\phi}{dt} \right| = \left| \frac{d}{dt}(\phi_0 - 4t) \right| = 4 V$.
Using the transformer ratio $V_s = V_p \left(\frac{N_s}{N_p}\right)$.
Secondary voltage $V_s = 4 \times \left(\frac{1500}{50}\right) = 4 \times 30 = 120 V$.
Detailed Solution (Best of Understanding Complex Concepts)
Concepts Involved
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Faraday’s Law of Electromagnetic Induction: This law states that the induced electromotive force (EMF or voltage) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit. Mathematically, the magnitude of the induced voltage $V$ is given by $V = \left\vert{} \frac{d\phi}{dt} \right\vert{}$.
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Transformer Principle: A transformer works on the principle of mutual induction. For an ideal transformer, the ratio of the voltage across the secondary coil ($V_s$) to the voltage across the primary coil ($V_p$) is equal to the ratio of the number of turns in the secondary coil ($N_s$) to the number of turns in the primary coil ($N_p$). This is expressed as $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Step-by-Step Solution
1. Identify the given information:
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Number of turns in the primary coil ($N_p$) = 50
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Number of turns in the secondary coil ($N_s$) = 1500
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Magnetic flux linked with the primary coil ($\phi$) = $\phi_0 - 4t$ (in Webers)
2. Calculate the voltage across the primary coil ($V_p$):
Using Faraday's Law, we differentiate the magnetic flux with respect to time to find the induced voltage in the primary coil.
$$V_p = \left\vert{} \frac{d\phi}{dt} \right\vert{}$$
$$V_p = \left\vert{} \frac{d}{dt}(\phi_0 - 4t) \right\vert{}$$
Since $\phi_0$ is a constant, its derivative is zero. The derivative of $-4t$ is $-4$.
$$V_p = \vert{}-4\vert{} = 4 \text{ Volts}$$
3. Calculate the output voltage across the secondary coil ($V_s$):
Using the transformer ratio equation, we can solve for the secondary voltage.
$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$
$$V_s = V_p \times \left( \frac{N_s}{N_p} \right)$$
Substitute the known values into the equation:
$$V_s = 4 \times \left( \frac{1500}{50} \right)$$
$$V_s = 4 \times 30$$
$$V_s = 120 \text{ Volts}$$
The output voltage across the secondary coil is 120 V.
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