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When a current of 5 mA is passed through galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the a be resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0-10 V is
To convert a galvanometer to a voltmeter, the series resistance (R) is calculated as: V = I(Rg + R) Where: V = 10 volts I = 5 mA = 0.005 A Rg = 15Ω 10 = 0.005 × (15+R) 15 + R = 10/0.005 = 2000 R = 2000 −15 = 1985Ω Answer: (C) 1.985 × 10³ Ω For more visit here: https://www.tiwariacademy.com/ncert-solRead more
To convert a galvanometer to a voltmeter, the series resistance (R) is calculated as:
V = I(Rg + R)
Where:
V = 10 volts
I = 5 mA = 0.005 A
Rg = 15Ω
10 = 0.005 × (15+R)
15 + R = 10/0.005 = 2000
R = 2000 −15 = 1985Ω
Answer: (C) 1.985 × 10³ Ω
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A galvanometer has a sensitivity of 60 division/ampere. When a shunt is used its sensitivity becomes 10 divisions/ampere. What is the value of shunt used if the resistance of the galvanometer is 20 Ω ?
Given: Initial sensitivity = 60 divisions/ampere Final sensitivity = 10 divisions/ampere Galvanometer resistance (G) = 20 Ω The relation is: S'/S = G/G + S 10/60 = 20/20+S 1/6 = 20/20+S 20 + S = 120 ⟹ S =100Ω The shunt value is: S = G×S′/S−S′ = 20×10/60−10 = 4Ω Answer: (C) 4 Ω For more visit here:Read more
Given:
Initial sensitivity = 60 divisions/ampere
Final sensitivity = 10 divisions/ampere
Galvanometer resistance (G) = 20 Ω
The relation is:
S’/S = G/G + S
10/60 = 20/20+S
1/6 = 20/20+S
20 + S = 120 ⟹ S =100Ω
The shunt value is:
S = G×S′/S−S′ = 20×10/60−10 = 4Ω
Answer: (C) 4 Ω
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A galvanometer of resistance 25 Ω is connected to a battery of 2 volt along with a resistance in series. When the value of this resistance is 3000 Ω, a full scale deflection of 30 units is obtained in the galvanometer. In order to reduce this deflection to 20 units, the resistance in series will be
Given: Galvanometer resistance (G) = 25 Ω Battery voltage (V) = 2 V Initial series resistance (R) = 3000 Ω Full-scale deflection = 30 units New deflection = 20 units The current sensitivity is proportional to deflection: I₂/I₁ = 20/30 = 2/3 Current through the circuit: I₁ = V/G+R = 2/3000+25 New resRead more
Given:
Galvanometer resistance (G) = 25 Ω
Battery voltage (V) = 2 V
Initial series resistance (R) = 3000 Ω
Full-scale deflection = 30 units
New deflection = 20 units
The current sensitivity is proportional to deflection:
I₂/I₁ = 20/30 = 2/3
Current through the circuit:
I₁ = V/G+R = 2/3000+25
New resistance:
I₂ = 2/G+R’
2/3/1 = G+R/G+R’
R’ = 4514Ω
Answer: (A) 4514 Ω
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A conducting wire of length l is turned In the form of a circular coil and a current l is passed through it. For the torque, due to magnetic field produced at its centre, to be maximum, the number of turns in the coil will be
The torque (τ) on a current-carrying coil is given by: τ = nI AB For fixed wire length l, the radius is: r = l/2πn Torque is maximized when n = 2, as the product of the number of turns and area is optimized. Answer: (B) Two For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/Read more
The torque (τ) on a current-carrying coil is given by:
τ = nI AB
For fixed wire length l, the radius is:
r = l/2πn
Torque is maximized when n = 2, as the product of the number of turns and area is optimized.
Answer: (B) Two
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The gyro-magnetic ratio of an electron in an H-atom, according to Bohr model, is
The gyro-magnetic ratio (γ) of an electron in a hydrogen atom, according to the Bohr model, is given by: γ = e/2m It is independent of the orbit number (n) as it only depends on fundamental constants (e and m). The ratio is negative because the electron has a negative charge. Answer: (B) Negative htRead more
The gyro-magnetic ratio (γ) of an electron in a hydrogen atom, according to the Bohr model, is given by:
γ = e/2m
It is independent of the orbit number (n) as it only depends on fundamental constants (e and m). The ratio is negative because the electron has a negative charge.
Answer: (B) Negative
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