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A series combination of two resistors 1 ...

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

Text Solution

Verified by Experts

The correct Answer is:
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`
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