NEET-XII-Physics
32: Electric Current in Conductors
- #17The potential difference between the terminals of a battery of emf 6.0 V and internal resistance 1 Ω drops to 5.8 V when connected across an external resistor. Find the resistance of the external resistor.Ans : Given:
Emf of the battery, E = 6 V
Internal resistance, r = 1 Ω
Potential difference, V = 5.8 V
Let R be the resistance of the external resistor.
Applying KVL in the above circuit, we get:
`` E\mathit{-}i\mathit{(}R\mathit{+}r\mathit{)}=0``
`` \Rightarrow i\mathit{=}\frac{E}{R\mathit{+}r}``
`` \mathit{=}\frac{6}{R\mathit{+}1}``
Also,
`` V\mathit{=}E\mathit{-}ir``
`` \Rightarrow 5.8=6-\frac{6}{R+1}\times 1``
`` \frac{6}{R+1}=0.2``
`` R+1=\frac{6}{0.2}=30``
`` R=29\,\mathrm{\,\Omega \,}``
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