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A reactive element, in an a.c. circuit, causes the current flowing (i) to lag in phase by π/2 , (ii) to lead in phase by π/2 with respect to the applied voltage. Identify the element in each case.
(i) The reactive element causing the current to lag the voltage by a phase angle of π/2 is an inductor. (ii) The reactive element causing the current to lead the voltage by π/2 is a capacitor. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
(i) The reactive element causing the current to lag the voltage by a phase angle of π/2 is an inductor.
(ii) The reactive element causing the current to lead the voltage by π/2 is a capacitor.
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A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere. You may take G = 6.67 x 10⁻¹¹ Nm²kg ⁻²
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitatioRead more
Given the mass of the sphere M = 100 kg, the mass of the particle m = 10 g or 10⁻² kg, and the radius R = 10 cm or 0.01 m, the initial potential energy (P.E.) of the two bodies is calculated based on their gravitational interaction. When the particle is moved far away from the sphere, the gravitational potential energy of the system becomes zero. The work done in moving the particle from its initial position to an infinite distance is the difference between the final and initial potential energies, resulting in a positive value.
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See lessThe instantaneous current and voltage of an a.c. circuit are given by I = 10 sin 300t A and V = 200 sin 300t V. What is the Power dissipation in the curcuit?
Since the current and voltage are in phase, the power factor cosϕ = 1. The power dissipation is given by P = V rms I rms For Vrms = 200V and I rms = 10 A, the power dissipation is P = 1/2 × 200 × 10 = 1000 W. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapterRead more
Since the current and voltage are in phase, the power factor cosϕ = 1. The power dissipation is given by P = V rms I rms
For Vrms = 200V and I rms = 10 A,
the power dissipation is P = 1/2 × 200 × 10 = 1000 W.
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The kinetic energy needed to project a body of mass m from the earth’s surface (radius R) to infinity is
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy rRead more
In the given scenario, the initial potential energy of the system is determined by the gravitational interaction between the two masses. When the particle is moved infinitely far away, the potential energy becomes zero. To move the particle from its initial position to infinity, the kinetic energy required is equivalent to the change in potential energy. This kinetic energy can be understood as the work needed to overcome the gravitational attraction between the two bodies. The amount of kinetic energy required is proportional to the gravitational acceleration, the mass of the particle, and the radius of the sphere.
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See lessEnergy required to move a body of mass m from an orbit of radius 2 R to 3 R is
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled,Read more
The gravitational potential energy (P.E.) of a mass m in an orbit of radius R is negative and proportional to the inverse of the radius. When the radius is doubled, the potential energy becomes less negative, reflecting the decrease in gravitational attraction. Similarly, when the radius is tripled, the potential energy further decreases. The change in gravitational potential energy between two orbits, such as from a radius of 2R to 3R, can be found by calculating the difference between the two potential energies. This difference represents the work done by the gravitational force in moving the object between these orbits.
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