Advertisements
Advertisements
प्रश्न
The power supply to the primary coil of a transformer is 200 W. Find
(i) Current in primary coil if the e.m.f. supply to it is equal to 220V.
(ii) The number of turns in the primary coil is equal to 80 and that in secondary is 800. What is the transformation ratio?
(iii) Name the type of transformer.
(iv) What will be the output voltage?
(v) What is the current in the secondary coil for an ideal transformer?
(vi) What is the output power?
(vii) Is output and input power equal?
(viii) Compare the current flowing in a secondary coil and in a primary coil.
Advertisements
उत्तर
(i) P(Power) = 200W = VI = Ep × Ip
∴ P = 200 = 200 × Ip
∴ Current in the primary coil = `200/220`
Ip = 0.909 A
(ii)
Transformation ratio `"E"_"s"/"E"_"p" = "N"_"s"/"N"_"p" = 800/80 = 10`
(iii) It is a step up transformer.
(iv) Using `"E"_"s"/"E"_"p" = "N"_"s"/"N"_"p" = 10`
` therefore "E"_"s"/200 = 10`
∴ Es = 2200 V
(v) Using Ep × Ip = 2200 × Is
∴ 200 = 2200 × Is
∴ Is = `2/22 = 0.090` A
(vi) Outpput power = Es × Is = `(2200 xx 0.0909)/100 = 200`W
(vii) Yes, It is equal, because it is an ideal transformer, for an ideal transformer
Efficiency = η = `"Outout power"/"Input power" = 1 "or" 100%`
i.e., there is no energy loss due to any cause.
(viii) Is = 0.0909 A and Ip = 0.909 A
Less current flows through the secondary coil than in the primary coil because it is a step-up transformer.
संबंधित प्रश्न
A group of students while coming from the school noticed a box marked "Danger H.T. 2200 V" at a substation in the main street. They did not understand the utility of a such a high voltage, while they argued, the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be an important question and therefore explained to the whole class.
Answer the following questions:
(i) What device is used to bring the high voltage down to low voltage of a.c. current and what is the principle of its working ?
(ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain
(iii) Write the values displayed by the students and the teacher.
The teachers of Geeta’s school took the students on a study trip to a power generating station, located nearly 200 km away from the city. The teacher explained that electrical energy is transmitted over such a long distance to their city, in the form of alternating current (ac) raised to a high voltage. At the receiving end in the city, the voltage is reduced to operate the devices. As a result, the power loss is reduced. Geeta listened to the teacher and asked questions about how the ac is converted to a higher or lower voltage.
1) Name the device used to change the alternating voltage to a higher or lower value. State one cause for power dissipation in this device.
2) Explain with an example, how power loss is reduced if the energy is transmitted over long distances as an alternating current rather than a direct current.
3) Write two values each shown by the teachers and Geeta.
Explain the significance of Lenz’s law to show the conservation of energy in electromagnetic induction.
Given the input current 15 A and the input voltage of 100 V for a step-up transformer having 90% efficiency, find the output power and the voltage in the secondary if the output current is 3 A.
The primary coil of a transformer has 200 turns while the secondary coil has 1000 turns. What type of transformer is this? if the input voltage is 10V, what will be the output voltage?
Devices which is used to convert high alternating current to low alternating current is ______.
State whether true or false. If false, correct the statement.
A transformer can step up direct current.
A transformer is essentially an a.c. device. It cannot work on d.c. It changes alternating voltages or currents. It does not affect the frequency of a.c. It is based on the phenomenon of mutual induction. A transformer essentially consists of two coils of insulated copper wire having different numbers of turns and wound on the same soft iron core.
The number of turns in the primary and secondary coils of an ideal transformer is 2000 and 50 respectively. The primary coil is connected to a main supply of 120 V and secondary coil is connected to a bulb of resistance 0.6 Ω.
The value of current in primary coil is ______.
A 12 V, 60 W lamp is connected to a 220 V AC supply using an ideal transformer. Find the primary current.
