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

Question

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 (A) 1 V (B) 2 V (C) 4 V (D) 6 V

Tardigrade Admin · Accepted Answer

i=(12 V /1 Ω+2 Ω+3 Ω)=(12 V /6 Ω)=2 A Potential difference across 2 Ω resistor =(2 A )(2 Ω)=4 V

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

1 V

2 V

4 V

6 V

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

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

1 V

2 V

4 V

6 V

i=1Ω+2Ω+3Ω12V​=6Ω12V​=2A Potential difference across 2Ω resistor =(2A)(2Ω)=4V

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

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