A series combination of two resistors `1 Omega` each is connected to a 12 V battery of internal resistance `0.4 Omega`. The current flowing through it

Question

A series combination of two resistors `1 Omega` each is connected to a 12 V battery of internal resistance `0.4 Omega`. The current flowing through it

A series combination of two resistors `1 Omega` each is connected to a 12 V battery of internal resistance `0.4 Omega`. The current flowing through it will be
A. 3.5A
B. 5A
C. 6A
D. 10A

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
Here, both the external resistance are connected in series combination of battery. Hence, total external resistance `= 1+1=2 Omega`
Hence, the toal resistance
= external resistance + internal resistance
`= 2+0.4=2.4 Omega`
So, current flowing is `l=(V)/(R )=(12)/(2.4)=5 A`

4 views Answered 2023-02-05 15:29