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Question
An 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.
Solution
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,
\[iR = i'R'\]
\[ \Rightarrow 5 \times R = 6 \times \frac{10R}{10 + R}\]
\[ \Rightarrow \left( 10 + R \right)5 = 60\]
\[ \Rightarrow 5R = 10\]
\[ \Rightarrow R = 2 \Omega\]
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