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Temperature dependence of resistivity p(T) of semiconductors, insulators and metals is significantly based on the following factors :
The temperature dependence of resistivity ρ(T) in semiconductors, insulators and metals is primarily influenced by: (a) Number of charge carriers can change with temperature T (especially in semiconductors and insulators). (b) Time interval between successive collisions depends on T (affecting electRead more
The temperature dependence of resistivity ρ(T) in semiconductors, insulators and metals is primarily influenced by:
(a) Number of charge carriers can change with temperature T (especially in semiconductors and insulators).
(b) Time interval between successive collisions depends on T (affecting electron scattering).
Thus, the correct answers are (a) and (b).
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A uniform wire of resistance 2 R is bent in the form of a circle. The effective resistance between the ends of any diameter of the circle is
When the wire of resistance 2 R is bent into a circle, each semicircular half has resistance R. These two halves are in parallel when measured across a diameter. The effective resistance is given by: 1/Reff = 1/R + 1/R Reff = R /2 Thus, the correct answer is (b) R/2. For more visit here: https://wwwRead more
When the wire of resistance 2
R is bent into a circle, each semicircular half has resistance
R. These two halves are in parallel when measured across a diameter. The
effective resistance is given by:
1/Reff = 1/R + 1/R
Reff = R /2
Thus, the correct answer is (b) R/2.
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See lessThree resistors having values R₁, R₂ and R₃ are connected in series to a battery. Suppose R₁ carries a current 2.0 A, R₂ has a resistance 3-0 ohm and R₃ dissipates 6.0 watt of power. Then voltage across R₃ is :
In a series circuit, the current remains the same across all resistors. Given that R₁ carries 2.0 A, R₃ also has 2.0 A. Power dissipated by R₃ is given by P = I² R. Solving 6 = (2)² R₃, we get R₃ = 1.5 ohm. Voltage across R₃ is V = IR = 2 × 1.5 = 3V. Answer: (c) 3 V. For more visit here: https://wwwRead more
In a series circuit, the current remains the same across all resistors. Given that
R₁ carries 2.0 A,
R₃ also has 2.0 A. Power dissipated by
R₃ is given by
P = I² R. Solving 6 = (2)² R₃, we get
R₃ = 1.5 ohm. Voltage across
R₃ is V = IR = 2 × 1.5 = 3V. Answer: (c) 3 V.
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A constant voltage is applied between the two ends of a uniform metallic wire, heat H is developed in it. If another wire of the same material, double the radius and twice the length as compared to original wire is used then the heat developed in it will be :
The heat developed in a wire is given by H = V²t / R. Resistance R is given by R = ρL/A.For the new wire: Length (L') = 2L, Radius (r') = 2r, so A' = 4A, R' = (ρ × 2L) / (4A) = R/2. Since H ∝ 1/R, the heat developed H' = 2H. Thus, the correct answer is (c) 2H. For more visit here: https://www.tiwariRead more
The heat developed in a wire is given by H = V²t / R. Resistance R is given by R = ρL/A.For the new wire:
Length (L’) = 2L,
Radius (r’) = 2r, so A’ = 4A,
R’ = (ρ × 2L) / (4A) = R/2.
Since H ∝ 1/R, the heat developed H’ = 2H.
Thus, the correct answer is (c) 2H.
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The current flowing through wire depends on time as, I = 3r + 2 t + 5. The charge flowing through the cross-section of wire in time t = 0 to t = 2 second is
To find the charge Q, integrate current I = 3t² + 2t + 5 over t = 0 to t =2: Q = ∫02(3t² +2t + 5)dt Solving, Q = t + t² + 5t 02 = (8+4+10) − (0) = 22C Answer: (a) 22 C. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
To find the charge
Q, integrate current I = 3t² + 2t + 5 over t = 0 to t =2:
Q = ∫02(3t² +2t + 5)dt
Solving,
Q = t + t² + 5t 02
= (8+4+10) − (0) = 22C
Answer: (a) 22 C.
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