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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?
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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
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