NEET-XII-Physics
32: Electric Current in Conductors
- #29An ideal battery sends a current of 5 A in a resistor. When another resistor of 10 Ω is connected in parallel, the current through the battery is increased to 6 A. Find the resistance of the first resistor.Ans : Let the resistance of the first resistor be R.
If V is the potential difference across R, then the current through it,
`` i=5=\frac{V}{R}``
Now, the other resistor of 10 Ω is connected in parallel with R.
It is given that the new value of current through the circuit, i' = 6 A
The effective resistance of the circuit, R' = `` \frac{10R}{10+R}``
Since the potential difference is constant,
`` iR=i\text{'}R\text{'}``
`` \Rightarrow 5\times R=6\times \frac{10R}{10+R}``
`` \Rightarrow \left(10+R\right)5=60``
`` \Rightarrow 5R=10``
`` \Rightarrow R=2\,\mathrm{\,\Omega \,}``
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