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.

Answer 1

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,
$$\begin{gathered} iR=i\text{'}R\text{'} \\ \Rightarrow 5\times R=6\times \frac{10R}{10+R} \\ \Rightarrow \left(10+R\right)5=60 \\ \Rightarrow 5R=10 \\ \Rightarrow R=2\mathrm{\Omega } \end{gathered}$$