Five identical resistors, each of value 1100Ω, are connected to a 220V battery as shown. The reading of ideal ammeter is : \[\frac{1}{3} A\]\[\frac{1}{5} A\]\[\frac{3}{5} A\]\[\frac{4}{5} A\]Solution:Current through each resistor is 220/1100 = 1/5 A Total current through ammeter = 3 * 1/5 A= 3/5 A
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