An ammeter of resistance 0.80Ω can measure currents upto 1 A . (i) What must be the shunt resistance to enable the ammeter to measure current upto 5A ? (ii) What is the combined resistance of the ammeter and the shunt?
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We have to make the current flowing in the internal resistance 0.80 Ω to 1/5 th of the current flowing ie, of 5 A.
When 1 A flows through the internal resistance 0.80 Ω, the remaining 4 Amp current flows through the shunt resistance S. Their currents are inversely proportional to their resistances, (they are in parallel)
so 0.80 Ω/ S = 4 A /1 A
S = 0.2 Ω = shunt resistance
Combined resistance of shunt and internal resistance: 0.8 * 0.2 / (0.8+0.2) = 0.16 Ohms
When 1 A flows through the internal resistance 0.80 Ω, the remaining 4 Amp current flows through the shunt resistance S. Their currents are inversely proportional to their resistances, (they are in parallel)
so 0.80 Ω/ S = 4 A /1 A
S = 0.2 Ω = shunt resistance
Combined resistance of shunt and internal resistance: 0.8 * 0.2 / (0.8+0.2) = 0.16 Ohms
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