Numerical on Transformer 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.
Numerical on Transformer 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 Transformer |
| Academic Session | 2025-2026 |
Numericals on Transformer
Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current
Que-61: A transformer is used to step up an alternating emf of 200 V to 440 V. If the primary coil has 1000 turns, calculate the number of turns in the secondary coil. (ISC 2019)
Ans- N1/N2 = V1/V2
=> 100/N2 = 200/440
=> N2 = 440 x 5 = 2200.
Que-62: For the given circuit comment on the type of transformer used.
Ans- Step-up transformer is used.
Que-63: A step-down transformer connected to an AC mains supply of 440 V is made to operate 22 V, 44 W lamp, ignore power losses in the transformer. Find the current in the primary circuit and transformation ratio.
Ans- Vp = 440 V , Vs = 22 V , Ps = 44 W
Ip = Pp/Vp
=> 44/440 = 0.1 A
r = Vs/Vp
=> 22/440 = 1/20
Que-64: A transformer has 1200 turns in the primary and 600 turns in the secondary winding. Primary voltage is 440 V and load in the secondary is 110 Ω. Calculate the current in the primary, assuming it to be an ideal transformer.
Ans- Vp/Vs = Np/Ns
=> Vs = Vp.Ns/Np
=> Vs = 440 x 600/1200 = 220 V
again, Is = Vs/RL
=> Is = 220 /110 = 2A
Now, Vp.Ip = Vs.Is
=> Ip = Vs.Is/Vp
=> 220 x 2 / 440 = 1 A
— : End of Numerical on Transformer Class-12 Nootan ISC Physics Solution Ch-12 Alternating Current :–
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