A milliameter with full scale deflection of 10mA has an internal resistance of 5 ohms. It would be converted to an ammeter with a full scale deflection of 1A by connecting a resistance of
\(\frac{5}{99}\)ohms in series with it
\(\frac{5}{99}\)ohms in parallel with it
\(\frac{99}{5}\)ohms in series with it
\(\frac{99}{5}\)ohms in parallel with it
2 ohms in series with it
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Discussions (4)
yeah it's B
to convert a miliameter (galvanometer) to read higher currents it's been connected in parallel with a low resistor(shunt) this is so that different current would flow through them and the sum of the current they detect is thus the total current flowing in the circuit
hmm never thought I could fail any question on this but I just did, thanks to my jotter
all the best
smarty
the voltage drop across the milliammeter and the the shunt connected is going to be the same
so Vm = Vs
(lR)m = (lR)s
All u need now is current of the shunt which is the difference between the required current and the milliammeter's current: 1 - 0.01(milli = exp -3) = 0.99
so Rs = (lR)m/ ls = 0.01*5/ 0.99 =
