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प्रश्न
A moving coil galvanometer has a coil of resistance 59 Ω. It shows a full-scale deflection for a current of 50 mA. How will you convert it to an ammeter having a range of 0 to 3A?
उत्तर
Let resistance of R, Ω be shunted with galvanometer to make it an ammeter in parallel.
p.d. across the galvanometer = p.d. across the shunt
IC1 x RC1 =(I- lC1 ) . RS
⇒ `R_s = (I_(C_1)R_(C_1))/(I-I_(C_1))=(50xx10^-3 xx59)/((3-50xx10^-3))`
⇒ `R_s = (50xx10^-3 xx59)/2.95 = (50xx10^-3 xx10^2 xx59)/295`
∴ `R_s = 1 Omega `
A resistance of 1 Ω be required to shunt in order to comvert this into ammeter.
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संबंधित प्रश्न
Write the underlying principle of a moving coil galvanometer.
An ideal voltmeter has _______.
(A) low resistance
(b) high resistance
(C) infinite resistance
(D) zero resistance
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of
- current sensitivity and
- voltage sensitivity of M2 and M1.
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How will you convert a moving coil galvanometer into a voltmeter?
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