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An ammeter measures currents by passing this the current to be measured through a mobile coil, and thus creating a proportional magnetic field. This field, in turn interacts with a second magnetic field created by a permanent magnet leading to a deflection of the indicator (mobile coil) proportional to the current passing through it. Because aways a coil has a resistance, there will be a voltage drop U onto this. The Ohm law states the relation between the current passing through the mobile coil I and its internal resistance R.
`I = U/R`
Therefore the full scale voltage using the problem data is
`U = I*R =500*10^-6 *400 = 0.2 "Volt"`
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