Question

The effective resistance between the terminals A and B is equal to:

(A) 25 ohms
(B) 15 ohms
(C) 10 ohms
(D) 5 ohms

Answer 1

To answer this question we need to know that series resistance is given by ${R_S} = {R_1} + {R_2} + {R_3}....... + {R_N}$, and parallel connection of resistances is given by $1/{R_p} = 1/{R_1} + 1/{R_2} + 1/{R_3}....... + 1/{R_N}$. Using this formula, we have to find out the equivalent resistance between the terminals A and B.

Complete step by step answer:
The circuit can be reduced, step by step, to a single equivalent resistance.
The 5 ohm and the 5 ohm that are connected to the terminal A are in series, and so they can be replaced by an equivalent resistor, which is denoted by ${R_{S1}}$. The value of ${R_{S1}}$ is 10 ohm, by using the equation given below:
${R_S}$= 5 + 5 = 10 ohms.
This is the resistance ${R_{S1}}$ of 10 ohms is now connected parallel to another 10 ohm resistance. This connection can be reduced to a single equivalent resistance ${R_p}$. They are connected in series and the equivalent resistance ${R_p}$is given as below:
$\dfrac{1}{{{R_p}}} = \dfrac{1}{{10}} + \dfrac{1}{{10}} = 5ohms$
Now the resistance ${R_p}$and 5 ohms are connected in series and the equivalent resistance ${R_{S2}}$ is given as ${R_{S2}}$= 5+5= 10 ohms.
The resistances ${R_{S1}}$and the remaining 10 ohm resistance are connected parallel and hence, the total resistance in the circuit is given as follows:
$\dfrac{1}{R} = \dfrac{1}{{10}} + \dfrac{1}{{{R_{S2}}}} = \dfrac{1}{{10}} + \dfrac{1}{{10}} = 5ohms$

Hence the resistance across the terminals A and B is 5 ohms. So the correct answer is option D.

Note: We should know that a parallel circuit will have two more than two paths for the current to pass through it. The sum of the currents passing through each of the paths will be equal to the total current that is flowing from the source. On the other hand, the total resistance in a series circuit will be equal to the sum of the individual resistances. Similarly, the voltage applied in the series circuit will be similar to the addition of all the individual voltage drops.