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In the circuit in figure, the current flowing `5 Omega` resistance is : A. 0.5 AB. 0.6 AC. 0.9 AD. 1.5 A |
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Answer» Correct Answer - B `I_(2)=(R_(1))/(R_(1)+R_(2))I=(10)/(35)xx2.1 =(21)/(35)=(3)/(5)` `I_(2) =0.6A`. |
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