The current in a single-loop circuit with one resistance Ris 5.0 A. When an additional resistance of 2.0$\mathbf{\Omega }$is inserted in series with R, the current drops to 4.0 A. What is R?

$$\begin{gathered} \mathrm{I}=5.0\mathrm{A} \\ {\mathrm{R}}^{\text{'}}=2.0\mathrm{\Omega } \\ {\mathrm{I}}^{\text{'}}=4.0\mathrm{A} \end{gathered}$$

Write for emf of the battery when the circuit has only resistance R and when R’ is inserted in the circuit using Ohm’s law. Equating them will give the value of R.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Where, I is current, V is voltage, R is resistance.

Let the battery be emf V. Then,Ohm’s law gives,

$$\begin{gathered} \mathrm{V}=\mathrm{IR} \\ \mathrm{V}=5\mathrm{R} \end{gathered}$$

When R’ is inserted in the circuit in series with R, the total resistance of the circuit becomes R+R’.

Then emf is given by,

$$\begin{gathered} \mathrm{V}={\mathrm{I}}^{\text{'}}\left(\mathrm{R}+{\mathrm{R}}^{\text{'}}\right) \\ \mathrm{V}=4\left(\mathrm{R}+2\right) \end{gathered}$$

Equating the equations 1) and 2) gives,

$$\begin{gathered} 5\mathrm{R}=4\mathrm{R}+8 \\ \mathrm{R}=8.0\mathrm{\Omega } \end{gathered}$$

Hence, the value of R is $8.0\mathrm{\Omega }$.

Therefore, using Ohm’s law, the unknown resistance in the circuit can be found.