Two resistance connected in series as shown in diagram. What is the value of $R$ ?
(A) $4\,\Omega $
(B) $6\,\Omega $
(C) $5\,\Omega $
(D) $3\,\Omega $

Hint: use the formula of the ohm’s law and rearrange it. Substitute the known potential difference and the resistance in it to find the current value. Use this current value in the ohm’s law again to find the value of the unknown resistance of the second resistor.
Useful formula:
(1) The formula of the ohm’s law is given by
$V = IR$
Where $V$ is the potential difference, $I$ is the current and $R$ is the resistance.

Complete step by step solution:
From the given diagram, the resistance of the first resistor, ${R_1} = 5\,\Omega $
Potential difference of the first resistor, ${V_1} = 10\,V$
Potential difference of the second resistor, ${V_2} = 6\,V$
The given circuit is the series circuit that consists of the two resistors of the known and the unknown resistors
Using the ohm’s law,
$V = IR$
By rearranging the formula, we get
$I = \dfrac{V}{R}$ ……………………(1)
Substitute the known parameters in the above step, we get
$I = \dfrac{{10}}{5} = 2\,A$
Substitute the condition for the second resistor in the equation (2),
$I = \dfrac{6}{{{R_2}}}$
Since this is a series circuit, the current in the resistor will be the same.
$
  2 = \dfrac{6}{{{R_2}}} \\
  {R_2} = 3\,\Omega \\
 $
Hence the resistance of the second resistor is $3\,\Omega $ .

Thus the option (D) is correct.

Note: Remember that if the resistors are connected in series, the current will be the same through each resistor and its potential difference will be equal to the sum of the potential difference developed in each resistor. If the resistor is connected in parallel, the current will be the sum of the current through each resistor and its potential difference will be the same to each resistor.