Question

Four resistances of 20 ohms each are connected in parallel. Four such combinations are connected in series. What is the total resistance?
A. $80$ ohms
B. $40$ ohms
C. $20$ ohms
D. $10$ ohms

Answer 1

Hint: In a parallel connection, the total resistance is equal to the reciprocal of the sum of the reciprocals of the individual resistances and four such combinations are then connected in series. In a series connection, the total resistance is equal to the sum of each resistance.

Complete step by step solution:
 In a parallel circuit, potential difference or voltage is the same across each end i.e., across each resistor and it is equal to the potential difference across the terminals of the battery or source. The current which is drawn from the source is the sum of current flowing through each branch of the parallel circuit.

We know that potential difference (V) is the product of current (I) and resistance (R) i.e., $V = IR\left[ {I = \dfrac{V}{R}} \right]$ and let ${I_1}$, ${I_2}$, ${I_3}$ and ${I_4}$ be the magnitude of current flowing through each parallel path. Hence,
$I$ =${I_1}$ +${I_2}$ + ${I_3}$ +${I_4}$
And $\dfrac{V}{R} = \dfrac{V}{{R_p}} + \dfrac{V}{{R_p}} + \dfrac{V}{{R_p}} + \dfrac{V}{{R_p}}$

Because the resistances connected in parallel are of the same value. Thus,
$\dfrac{V}{R} = \dfrac{{4V}}{{R_p}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{4}{{R_p}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{4}{{20}}$
$\Rightarrow R = 5$ ohms

The total resistance across each parallel branch is 5 ohms and four such combinations i.e., 5 ohms are connected in series. In a series combination, the current flowing through each resistor is the same because the current has a single path for its flow and potential difference across the entire circuit is equal to the sum of potential differences across the individual resistor. Let ${V_1}$, ${V_2}$, ${V_3}$ and ${V_4}$ be the potential differences across each resistor. Hence,
${V_s} ={V_1} + {V_2} + {V_3} +{V_4}$ where ${V_s}$ is the potential difference across the entire circuit.
$I{R_s} = IR + IR + IR + IR$ where ${R_s}$ is the total resistance
$\Rightarrow I{R_s} = 4IR$
$\Rightarrow{R_s} = 4R$
$\Rightarrow{R_s} = 4 \times 5$
$\therefore{R_s} = 20$ ohms

Therefore, option C is the correct answer.

Note: Initially four resistances are connected in parallel and then four resistances with equal value i.e., equal to the value of resultant resistance of the parallel connection are connected in series. So, kindly read the question properly which would help to solve it easily. Also, remember that the net resistance in parallel connection is the reciprocal of the sum of the reciprocals of the individual resistances and it is the sum of each resistance in series connection.