V I L-C Circuits and Oscillations Numerical Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
V I L-C Circuits and Oscillations Numerical Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-12 | Alternating Current |
| Topics | Numericals on V I L-C Circuits and Oscillations |
| Academic Session | 2025-2026 |
Numericals on V I L-C Circuits and Oscillations
Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current
Que-47: In an L-C circuit connected to an AC source the potential difference across L is 50 V and that across C is 30 V. Find the voltage of the source.
Ans- V = I VL – VC I
=> V = I 50 – 30 I =20 V
Que-48: A 10 µF capacitor and a 1.0 H inductor are connected in series with an AC source of frequency 50 Hz. Find the impedance of the circuit.
Ans- Z = I XL-XC I
=> I 100π – 10³/π I
=> I -4.47 I
=> 4.47 Ω
Que-49: A 200 km long telegraph wire has a capacitance of 0.014 µF/km. If it carries an AC of 5 kHz, what should be the inductance required to be connected in series so that the impedance is minimum?
Ans- L – 1 / (2π.5 x 10³ Hz)² . 2.8 x 10^-6 F
=> 1 / (4π².25 x 10^6 Hz². 2.8 x 10^-6 F) = 0.36 x 10^-3 H
L = 0.36 mH
Que-50: An AC source of frequency 50 Hz is connected to a 50 mH inductor and a bulb. The bulb glows with some brightness. Calculate the capacitance of the capacitor to be connected in series with the circuit, so that the bulb glows with maximum brightness.
Ans- XL = 2πfL
=> 2π(50)(50 x 10^-3) = 5π Ω
C = 1/2πf XL
=> 1/2π(50).(5π)
=> 2.03 x 10^-4 F
Que-51: Find the resonant angular frequency of a series L-C-R What is circuit with L = 2.0 H , C = 32 µF and R = 10 Ω . What is the Q-value of this circuit?
Ans- Q = 1/R √L/C
=> 1/10 √1/(16 x 10^-3) = 25
for resonant frequency XL = XC
=> ωL = 1/ωC = ω = 1/√LC
=> 1 / √(2 x 32 x 10^-6)
=> 1000/8 = 125 rad/s
— : End of V I L-C Circuits and Oscillations Numerical Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current . :–
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