Three resistors of 2 Ω, 4 Ω, and 6 Ω are connected in series across a 12 V battery. The current in the circuit is:
Solution
$R_{eq} = 2+4+6 = 12$ Ω. $I = V/R = 12/12 = 1$ A
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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