AC Circuit with Either of R L or C 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.
AC Circuit with Either of R L or C 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 AC Circuit with Either of R L or C |
| Academic Session | 2025-2026 |
Numericals on AC Circuit with Either of R L or C
Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current
Que-7: What is the self-inductance of a coil if its inductive reactance is 12.56 Ω at a frequency of 200 Hz?
Ans- XL = ωl
=> L = XL/ω = XL/2πf
=> 12.56 / 2 x 3.14 x 200 = 0.01 H
Que-8: At what frequency will a 2 H inductor have a reactance of 4000 Ω?
Ans- XL = ωl = 2πfL
=> f = XL/2πf
=> 4000 / 2 x π x 2 = 1000/π Hz
Que-9: A pure inductor of 0.5 H carries an AC of 0.2 A and of frequency 50 Hz. What will be the voltage across the ends of the inductor and the phase difference between the voltage and the current?
Ans- L = 0.5 H , f = 50 Hz , i = 0.2 A
XL = ωl = 2πfL
=> 2 x 3.14 x 50 x 0.5
VL = i x XL
=> 0.2 x 2 x 3 .14 x 50 x 0.5 = 31.4 volt
In case of pure inductive circuit voltage leads current by 90°
Que-10: A pure inductance of 1.0 H is connected across a 110 V-70 Hz source. Find reactance, current and peak value of current.
Ans- XL = ωl = 2πfL
=> 2 x 22/7 x 70 x 1 = 440 Ω
Current = V/XL = 110/440 = 0.25 A
and peak value of current
=> 0.25 x √2 = 0.35 A
Que-11: A capacitor of 50 pF is connected to an AC source of frequency 1 kHz. Calculate its reactance.
Ans- C = 50 x 10^-12 F
f = 1 x 10³ Hz
Xc = 1/ωc = 1/2πfc
=> 1/2π x 10³ x 50 x 10^-12
=> 10^7/π Ω
Que-12: A 2.0 µF capacitor is connected to a 220 V, 50 Hz AC source. Calculate the reactance of the capacitor, the current flowing in the circuit and mention the phase relationship between the current and the voltage.
Ans- C = 2 x 10^-6 F , V = 220 volt , f = 50 Hz
XC = 1/ωC = 1/2πfC
=> 1 / 2 x 3.14 x 50 x 2 x 10^-6 = 1592.3 Ω
i = V/XC = 220/1592.3 = 0.138 A
In case of pure capacitive circuit current leads voltage by 90°
Que-13: Find the frequency at which the capacitive reactance of a 1.0 µF capacitor equals the inductive reactance of a 1.0 H coil.
Ans- XL = XC = ωL = 1/ωC
=> ω = 1/√LC => 2πf 1/√LC
f = 10³ / 2 x 3.14 = 159.24 Hz
Que-14: A coil has an inductance of 1 H. At what frequency will it have a reactance of 3140 Ω? What capacitance will have the same reactance at that frequency?
Ans- XL = ωl = 2πfL
=> f = XL/2πL
=> 3140 / 2 x 3.14 x 1 = 500 Hz
for capacitance XC = 1/ωC => C = 1/XC x ω
=> 1 / 3140 x 2 x 3.14 x 500 = 1 x 10^-7 F
— : End of AC Circuit with Either of R L or C Numerical Class-12 Nootan ISC Physics Solution Ch-12 :–
Return to : – Nootan Solutions for ISC Class-12 Physics
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