The output voltage of a transformer connected to a 220 V line is 1100 V at 2 A current. Its efficiency is 100%. The current coming from the line is:
1
20 A
2
10 A
3
1 A
4
2 A
Official Solution
Correct Option:
(2)
Step 1: Use the power formula. The power in the primary coil is equal to the power in the secondary coil (since efficiency is 100%).
Thus, for the secondary coil: Step 2: Calculate the current in the primary coil. Using the power equation for the primary coil:
Solving for , we get . Final Answer:
02
PYQ 2012
medium
physicsID: viteee-2
In an AC circuit, the potential across an inductance and resistance joined in series are respectively 16V and 20V. The total potential difference across the circuit is
1
200V
2
256V
3
319V
4
336V
Official Solution
Correct Option:
(3)
In an AC circuit, the total potential difference is the vector sum of the potential differences across the inductor and resistor. Using the Pythagorean theorem, we calculate the total potential difference.
Step 2: Conclusion.
The total potential difference across the circuit is 319V, corresponding to option (c).
03
PYQ 2012
medium
physicsID: viteee-2
Two capacitors of capacitance and are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacitance is
1
1800
2
1600
3
1400
4
7200
Official Solution
Correct Option:
(2)
For capacitors connected in series, the same charge is stored on both capacitors. The energy stored in a capacitor is given by , where is the capacitance and is the potential difference across the capacitor. We calculate the energy stored in the first capacitor.
Step 2: Conclusion.
The energy stored in the capacitor of capacitance is 1600 , corresponding to option (b).
04
PYQ 2012
medium
physicsID: viteee-2
A parallel plate capacitor has capacitance . If it is equally filled the parallel layers of materials of dielectric constant and , its capacity becomes . The ratio of and is
1
2
3
4
Official Solution
Correct Option:
(3)
For a parallel plate capacitor filled with two dielectric materials, the effective dielectric constant is given by the formula , considering the properties of the materials.
Step 2: Conclusion.
The correct ratio is , corresponding to option (c).
05
PYQ 2012
medium
physicsID: viteee-2
A resistor , an inductor , and a capacitor are connected in series to an oscillator of frequency . If the resonant frequency is , then the current lags behind voltage when
1
2
3
4
Official Solution
Correct Option:
(2)
At resonance, the impedance of the circuit is minimized, and the voltage and current are in phase. When , the circuit behaves inductively, and the current lags behind the voltage.
Step 2: Conclusion.
The current lags behind the voltage when , corresponding to option (b).
06
PYQ 2013
medium
physicsID: viteee-2
The capacitance of a parallel plate capacitor with air as dielectric is . If a slab of dielectric constant and of the same thickness as the separation between the plates is introduced so as to fill th of the capacitor (shown in figure), then the new capacitance is
1
2
3
4
None of these
Official Solution
Correct Option:
(2)
Step 1: Capacitance with dielectric.
When a dielectric slab is introduced into a capacitor, the capacitance increases by a factor depending on the dielectric constant. The new capacitance is calculated based on the volume fraction of the dielectric inserted into the capacitor.
Step 2: Conclusion.
The new capacitance is , which corresponds to option (2).
07
PYQ 2015
medium
physicsID: viteee-2
The value of inductance for which the current is maximum in series circuit with and is
1
10 mH
2
50 mH
3
200 mH
4
100 mH
Official Solution
Correct Option:
(3)
The resonance condition for a series circuit is given by , where is the resonant angular frequency. Using the given values for and , we can calculate the inductance .
About Ac Circuits - VITEEE
Ac Circuits is a vital chapter for VITEEE aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.
By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.
Frequently Asked Questions
Why focus on Ac Circuits PYQs?
Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.
How to best use this analysis?
Review the topic breakdown to see which sub-topics within Ac Circuits carry the most weight. Then, tackle the questions iteratively to solidify your understanding.
