Question

For the given circuit, calculate
(i) equivalent resistance between the points A and C.
(ii) current in the circuit
(iii) potential difference between A and B

Answer 1

(i) The resistance across AB and resistance across BC is in series and the equivalent resistance of these two will be parallel to the $2\Omega$ resistance.
(ii) Find out the equivalent resistance of the circuit. Current across any resistance $R$ is given by $I = \dfrac{V}{R}$ where $V$ is the potential difference across the resistance.
(iii) Voltage or potential difference across resistors in parallel connection is same and current across resistors in series connection is same. Find the potential difference across AC and then current through ABC. Finally, the potential difference between A and B will be ${V_{AB}} = I{R_{AB}}$ .

Complete step by step answer:
(i) As clear from the figure that both the $1\Omega$ resistances are connected in series and we know that if two resistance are connected in series then their equivalent resistance is the algebraic sum of the two resistances. So, the equivalent resistance of the $1\Omega$ resistances will be ${R_1} = 1\Omega + 1\Omega = 2\Omega$ .
Now, ${R_1}$ will be parallel to the $2\Omega$ resistance. So, the equivalent resistance across AC will be
${R_{AC}} = \dfrac{{2 \times 2}}{{2 + 2}} = 1\Omega$
Hence, the equivalent resistance between the points A and C is $1\Omega$ .
(ii) To find the current in the circuit, we have to first calculate the equivalent resistance of the circuit.
The equivalent resistance of the circuit is $1\Omega$ .
We know that the current across any resistance $R$ is given by $I = \dfrac{V}{R}$ where $V$ is the potential difference across the resistance.
So, the current through the circuit
$I = \dfrac{2}{1} = 2A$ .
(iii) As clear from the figure that both the $1\Omega$ resistances are connected in series and they are parallel with the $2\Omega$ resistance. We know that the voltage or potential difference across resistors in parallel connection is the same and current across resistors in series connection is the same.
So, the potential difference across the equivalent resistance ${R_1} = 2\Omega$ is $V = 2V$ .
Therefore, current through ABC will be $I = \dfrac{2}{2} = 1\Omega$ .
Now, the potential difference between A and B will be ${V_{AB}} = I{R_{AB}}$
So, ${V_{AB}} = 1 \times 1 = 1V$
Hence, the potential difference between A and B is $1V$ .

Note: The relation between current and voltage is explained by the Ohm’s Law which states that the amount of electric current passing through a conductor in a circuit is directly proportional to the voltage or potential difference across it, at a fixed temperature and similar physical conditions.