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,

\[iR = i'R'\]

\[ \Rightarrow 5 \times R = 6 \times \frac{10R}{10 + R}\]

\[ \Rightarrow \left( 10 + R \right)5 = 60\]

\[ \Rightarrow 5R = 10\]

\[ \Rightarrow R = 2  \Omega\]