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  4. Three resistors 1 Ω, 2 Ω, 3 Ω are combined in series. They are connected across a 12 V battery of negligible internal resistance. The potential drop across the 2 Ω resistor is

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Q. Three resistors $1 \Omega, 2 \Omega, 3 \Omega$ are combined in series. They are connected across a $12 V$ battery of negligible internal resistance. The potential drop across the $2 \Omega$ resistor is

Current Electricity

Solution:

$i=\frac{12 V }{1 \Omega+2 \Omega+3 \Omega}=\frac{12 V }{6 \Omega}=2 A$
Potential difference across $2 \Omega$ resistor
$=(2 A )(2 \Omega)=4 V$