In a balanced condition, resistance R3 has temperature coefficient 0.0004 degree C−1

As per the question, resistance R3 has a temperature coefficient of 0.0004^∘C⁻¹in a balanced condition. And, the temperature of R3 is increased by 100^∘C. In order to determine the voltage generated between S and T, we have to consider the following formula:

R¹= R₀ (1 + \(\alpha\)\(\Delta\)T)

Values of Resistance

Values of Resistance

Thus, after replacing the values, we get:

⇒ 300 (1 + 0.0004 \(\times\) 100)

Therefore, R¹ = 312 \(\Omega\)

I₁ = \({{50} \over 370}\)A

I₂ = \({{50} \over 600}\)A

⇒ V_s – \({{50} \over 370}\) \(\times\) 312 + \({{50} \over 600}\) \(\times\) 500 = V_T

V_s– V_T = 0.26 Volts (V)

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In a balanced condition, resistance R3 has temperature coefficient 0.0004 degree C 1. With temperature of R3 increased by 100 degree C, find voltage generated between S and T.

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A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

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In a balanced condition, resistance R3 has temperature coefficient 0.0004 degree C 1. With temperature of R3 increased by 100 degree C, find voltage generated between S and T.

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CBSE CLASS XII Related Questions

1.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

2.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

5. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.

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Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

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A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

3.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

4.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

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6.
Assertion (A) : All atoms have a net magnetic moment. Reason (R) : A current loop does not always behave as a magnetic dipole.