In a balanced Wheatstone bridge, the central galvanometer branch can be ignored. The two parallel arms each have resistance 2R, and their equivalent resistance is: Req = 2R×2R/2R +2R = R Answer: (a) R For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
In a balanced Wheatstone bridge, the central galvanometer branch can be ignored.
The two parallel arms each have resistance 2R, and their equivalent resistance is:
Using the meter bridge formula: R/S = L/100 - L Given R = 2Ω and L = 40 cm: 2/S = 40/60 S = 2 × 60/40 = 3Ω Answer: (b) 0.5 Ω For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Using the meter bridge formula:
R/S = L/100 – L
Given R = 2Ω and
L = 40 cm:
2/S = 40/60
S = 2 × 60/40 = 3Ω
Answer: (b) 0.5 Ω
Drift velocity vd is given by = vd = eEτ/m, Where E = V/L. Thus, vd = eVr/mL Since vd is directly proportional to V, the correct answer is (c) V. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Drift velocity vd is given by = vd = eEτ/m, Where E = V/L. Thus,
vd = eVr/mL
Since vd is directly proportional to V, the correct answer is (c) V.
Current is given by I = Q/t For A: I = 300/60 = 5A (correct). For B: Charge Q = (3.125 10) (1.6 10) = 5C, so I = 5A (correct). Thus, both A and B are correct. Answer: (c) both A and B. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Current is given by I = Q/t
For A: I = 300/60 = 5A (correct).
For B: Charge Q = (3.125 10) (1.6 10) = 5C, so I = 5A (correct).
Thus, both A and B are correct. Answer: (c) both A and B.
The resistance of the bulb is R =220²/100= 484Ω . When operated at 110V, power consumed is P ′ = 110²/484 = 25W. thus, the Correct answer: (a) 25 W. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
The resistance of the bulb is
R =220²/100= 484Ω
. When operated at 110V, power consumed is
P ′ = 110²/484 = 25W.
thus, the Correct answer: (a) 25 W.
In Wheatstone’s bridge, all the four arms have equal resistance R. If resistance of the galvanometer arm is also R, then equivalent resistance of the combination is
In a balanced Wheatstone bridge, the central galvanometer branch can be ignored. The two parallel arms each have resistance 2R, and their equivalent resistance is: Req = 2R×2R/2R +2R = R Answer: (a) R For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
In a balanced Wheatstone bridge, the central galvanometer branch can be ignored.
The two parallel arms each have resistance 2R, and their equivalent resistance is:
Req = 2R×2R/2R +2R = R
Answer: (a) R
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For a cell of emf 2 V, a balance is obtained for 50 cm of the potentiometer wire. If the cell is shunted by a 2 n resistor and the balance is obtained across 40 cm of the wire, then the internal resistance of the cell is:
Using the meter bridge formula: R/S = L/100 - L Given R = 2Ω and L = 40 cm: 2/S = 40/60 S = 2 × 60/40 = 3Ω Answer: (b) 0.5 Ω For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Using the meter bridge formula:
R/S = L/100 – L
Given R = 2Ω and
L = 40 cm:
2/S = 40/60
S = 2 × 60/40 = 3Ω
Answer: (b) 0.5 Ω
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
When a potential difference V is applied across a conductor at temperature T, the drift velocity of the electrons is proportional to :
Drift velocity vd is given by = vd = eEτ/m, Where E = V/L. Thus, vd = eVr/mL Since vd is directly proportional to V, the correct answer is (c) V. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Drift velocity vd is given by = vd = eEτ/m, Where E = V/L. Thus,
vd = eVr/mL
Since vd is directly proportional to V, the correct answer is (c) V.
For more visit here:
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Two students A and B calculate the charge flowing through a circuit. A concludes that 300 C of charge flows in 1 minute. B concludes that 3.125 x 10¹⁹ electrons flow in 1 second. If the current measured in the circuit is 5 A, then the correct calculation is done by
Current is given by I = Q/t For A: I = 300/60 = 5A (correct). For B: Charge Q = (3.125 10) (1.6 10) = 5C, so I = 5A (correct). Thus, both A and B are correct. Answer: (c) both A and B. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
Current is given by I = Q/t
For A: I = 300/60 = 5A (correct).
For B: Charge Q = (3.125 10) (1.6 10) = 5C, so I = 5A (correct).
Thus, both A and B are correct. Answer: (c) both A and B.
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The electric power consumed by a 220 V-100 W bulb when operated at 110 V is:
The resistance of the bulb is R =220²/100= 484Ω . When operated at 110V, power consumed is P ′ = 110²/484 = 25W. thus, the Correct answer: (a) 25 W. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
The resistance of the bulb is
R =220²/100= 484Ω
. When operated at 110V, power consumed is
P ′ = 110²/484 = 25W.
thus, the Correct answer: (a) 25 W.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/