Write the equivalent resistance of the mentioned circuit.

Hint: When we look into the circuit diagram we can see that the resistances are connected in a series manner. When three resistors are connected in series, then their equivalent resistance will be the sum of the individual resistance values. Substitute the values in the equation and find out the equivalent resistance. This will help you in answering this question.

Complete answer:
When we look into the circuit diagram we can see that the resistances are connected in a series manner. That is, the end of a resistor has been directly connected with the other resistor using a wire. When three resistors are connected in series, then their equivalent resistance will be the sum of the individual resistance values. Therefore the series connection will be providing the effective resistance which can be written as,
${{R}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}$
Where ${{R}_{eq}}$ be the equivalent resistance of the circuit, ${{R}_{1}}$ be the resistance of the first connected resistor which has been mentioned in the question as,
${{R}_{1}}=1\Omega $
${{R}_{2}}$ be the resistance of the second connected resistor which has been given as,
\[{{R}_{2}}=2\Omega \]
${{R}_{3}}$ be the resistance of the third resistor which has been mentioned as,
${{R}_{3}}=3\Omega $
Substituting these values in the equation will give,
\[{{R}_{eq}}=1+2+3=6\Omega \]
Therefore the effective resistance of the circuit has been found to be as \[6\Omega \].
The answer for the question has been calculated.

Note:
In the case of the series connection, the current which is flowing through each of the components will be similar, and the voltage across the circuit will be the addition of the individual voltage drops across each component. In an opposite way, in a parallel circuit, the voltage across each of the components will be similar, and the total current will be the resultant of the currents flowing through each component.