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A bulb of resistance 5Omega is connected...

A bulb of resistance `5Omega` is connected to a battery of emf `4.2` V and internal resistance `1Omega`. Find the current through the bulb.

Text Solution

Verified by Experts

`I=(E)/(R+r)=(4.2V)/(5Omega+1Omega)`
`=(4.2V)/(6Omega)=0.7A`
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