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A metal rod of length 2m rotates vertically about one of its end with frequency 2 Hz. The horizontal component of earth's magnetic field is $3.14 \times 10^{-5} T$ then emf developed between two ends of road is.....
The self inductance of a coil is 5H, a current of 1A changes to 2Awithin 5 sec. through the coil. The value of induced emf will be.......
The induced electromotive force (emf) in a coil is given by Faraday's Law of Electromagnetic Induction, which states that:
\[ \text{emf} = -L \frac{dI}{dt} \]
Where:
Here, the current changes from 1A to 2A in 5 seconds. Therefore, \(\frac{dI}{dt} = \frac{2A - 1A}{5s} = \frac{1}{5} = 0.2 A/s\).
Substituting the values, we get:
\[ \text{emf} = -5H \times 0.2 A/s = -1 V \]
The negative sign indicates the direction of the induced emf. Thus, the magnitude of the induced emf is 1V.
A coil of inductance 300 mH and resistance $ 2 \Omega $ is connected to a source of voltage 2V. The current reaches half of its steady state value in.........
If rotational velocity of a dynamo armature is doubled, then induced emf will become...What is increased in step down transformer ?
The induced emf (electromotive force) in a dynamo is directly proportional to the rotational velocity of the armature. According to Faraday's Law:
\[ \text{emf} \propto \text{rotational velocity} \]
If the rotational velocity of the dynamo armature is doubled, the induced emf will also double. Therefore, if the original emf is \(E\), then the new emf when the rotational velocity is doubled will be \(2E\). Thus, the induced emf will become two times the original value.
The core of a transformer is laminated so that.......
fact
In transformer, core is made of soft iron to reduce.....
A primary winding of transformer has 500 turns whereas its secondary has 5000 turns. Primary is connected to ac supply of 20V, 50Hz The secondary output of....
The transformer works on the principle of mutual induction and the voltage transformation ratio is given by the formula: \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). Here, \(V_p\) is the primary voltage, \(V_s\) is the secondary voltage, \(N_p\) is the number of primary turns, and \(N_s\) is the number of secondary turns. Given \(V_p = 20V\), \(N_p = 500\), and \(N_s = 5000\), the secondary voltage \(V_s\) can be calculated as: \[ V_s = V_p \times \frac{N_s}{N_p} = 20V \times \frac{5000}{500} = 200V \] The frequency remains the same at 50 Hz. Hence, the correct answer is
200V, 50Hz
.A step down transformer is connected to main supply 200 V to operate a 6V, 30 w bulb. The current in primary is.....
To find the current in the primary coil, we first need to determine the current in the secondary coil. The power consumed by the bulb is 30W and the voltage across it is 6V. Using the formula \( P = V \times I \), we get: \[ I_s = \frac{P}{V} = \frac{30W}{6V} = 5A \] The transformer is step-down, so the primary voltage is higher than the secondary. The current transformation ratio is given by \( \frac{I_s}{I_p} = \frac{V_p}{V_s} \). Given \(V_p = 200V\) and \(V_s = 6V\), we can calculate the primary current \(I_p\) as: \[ I_p = I_s \times \frac{V_s}{V_p} = 5A \times \frac{6V}{200V} = 0.15A \] Hence, the current in the primary is 0.15 A.
An emf of 15 V is applied in a circuit containing 5H inductance & $ 10 \Omega $ resistance. The ratio of the currents at time $t = \infty $ and at t = 1sec
$ I = I_0 (1 - e^{-Rt \over L}) $ $ I_\alpha = I _0 (1 -e ^{- \alpha} ) = 1.5 $ $ I1 = 1.5 (1- e -2) \Rightarrow { I_\alpha \over I_1 } = { 1 \over 1 -e^{-2} } $
Two coils have a mutual inductance 0.005H The current changes in a coil according to equation $ I =I_0 sin \omega t $ where $I_0 = 10 A$ and $\omega = 100 \pi rad s^{-1} $. The maximum value of emf in second coil is
$ \varepsilon = M { dI \over dt} \Rightarrow 0.005 { d \over dt } ( 10 sin 100 \pi t ) = \varepsilon _ { max } = 5 \pi $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.