A galvanometer has a resistance of 100 Ω. A current of 10-3 A pass through the galvanometer How can it be converted

Q: A galvanometer has a resistance of 100 Ω. A current of 10-3 A pass through the galvanometer How can it be converted into (a) ammeter of range 10 A and (b) voltmeter of range 10 volt .

Sol: G = 100 Ω ; Ig = 10-3 A

(a) I = 10 A; N = I/Ig = 104

S = G/((n-1)) = 100/((104-1)) = 100/999 Ω

(b) Vg = Ig G = 10-3 × 100 = 10-1 V

V = 10 V

⇒ n = V/Vg = 10/(10-1 ) = 100

∴ R = G(n – 1) = 100(100 – 1) = 9900 Ω

A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20 Ω, Calculate the resistance to be connected…

Q: A maximum current of 0.5 mA can be passed through a galvanometer of resistance 20 Ω, Calculate the resistance to be connected is series to convert it into a voltmeter of range (0 – 5)V.

Sol: R = G(n – 1), where n = v/Vg

V = 5V ; Vg = ig G = 0.5 × 10-3 × 20 = 10-2 V

∴ n = 500 and R = 20(500 – 1) = 9980 Ω

A galvanometer has a resistance of 98 Ω. If 2% of the main current is to be passed through the meter, what should be the value of the shunt ?

Q: A galvanometer has a resistance of 98 Ω. If 2% of the main current is to be passed through the meter, what should be the value of the shunt?

Sol: G = 98 Ω ; $\large \frac{i_g}{i} \times 100 = 2 \%$

$\large i_g G = i_s S$

$\large S = \frac{i_g G}{i_s} = \frac{i_g G}{i – i_g}$

$\large S = \frac{G}{i/i_g – 1}= \frac{98}{50-1}$

S = 2 ohm

The resistance of galvanometer is 999 Ω. A shunt of 1 Ω is connected to it. If the main current is 10^(-2) A…

Q: The resistance of galvanometer is 999 Ω. A shunt of 1 Ω is connected to it. If the main current is 10-2 A, what is the current flowing through the galvanometer.

Sol: G = 999 Ω , S = 1 Ω ,  i = 10-2 A ; ig = ?

$\large i_g = \frac{S}{G+S}i$

$\large i_g = \frac{1}{99+1} \times 10^{-2}$

= 10-5 A

A galvanometer has resistance 500 ohm. It is shunted so that its sensitivity decreases by 100 times. Find the shunt resistance.

Q: A galvanometer has resistance 500 ohm. It is shunted so that its sensitivity decreases by 100 times. Find the shunt resistance.

Sol: $\large Sensitivity \propto \frac{1}{Range}$

n = 100 ;

$\large S = \frac{G}{n-1} = \frac{500}{100-1}$

$\large S = \frac{500}{99}$

S = 5.05 ohm

A galvanometer of resistance 20 Ω is shunted by a 2 Ω resistor. What part of the main current flows through the galvanometer ?

Q: A galvanometer of resistance 20 Ω is shunted by a 2 Ω resistor. What part of the main current flows through the galvanometer ?

Sol: Ig G = Is S

Ig G = ( I – Ig ) S

Ig ( G + S ) = I S

Given G = 20 Ω ; S = 2 Ω

$\large \frac{I_g}{I} = \frac{S}{G + S} = \frac{2}{20 + 2}$

$\large \frac{I_g}{I} = \frac{1}{11}$

Hence , (1/ 11)th part of current is passing through galvanometer.

A coil area 100 cm^2 having 500 turns carries a current of 1 mA. It is suspended in a uniform magnetic field of induction…

Q: A coil area 100 cm2 having 500 turns carries a current of 1 mA. It is suspended in a uniform magnetic field of induction 10-3 Wb/m2 . Its plane makes an angle of 60° with the lines of induction. Find the torque acting on the coil.

Sol: Given = 1mA = 10-3 A ; N = 500 ; B = 10-3 Wb/m2 θ = 60°,

τ = ? A = 100 cm2 = 100 × 10-4 m2

Couple acting on the coil is given by

τ = BiAN sin α  ; Where α is angle made by normal to the plane of coil with B .

⇒ α = 90 – 60 = 30°

∴ τ = 10-3 × 10-3 × 100 × 10-4 × 500 × sin 30

= 250 × 10-8 Nm.