Consider the following part of an electric circuit : [Image of circuit

Consider the following part of an electric circuit :
[Image of circuit diagram is implied here]
The total electrical resistance in the given part of the electric circuit is

$ rac{15}{8}$ ohm
$ rac{15}{7}$ ohm
15 ohm
$ rac{17}{3}$ ohm
This question was previously asked in
UPSC NDA-2 – 2019
While the circuit diagram is not provided in the text, assuming a common configuration that results in one of the given simple fractional values, let’s test a common parallel combination. If the circuit consists of two resistors with resistances 3 ohm and 5 ohm connected in parallel, the total electrical resistance ($R_{total}$) is given by the formula for parallel resistors:
$R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}$
Let $R_1 = 3 \, \Omega$ and $R_2 = 5 \, \Omega$.
$R_{total} = \frac{3 \, \Omega \times 5 \, \Omega}{3 \, \Omega + 5 \, \Omega} = \frac{15 \, \Omega^2}{8 \, \Omega} = \frac{15}{8} \, \Omega$.
This result matches option A. This configuration is a plausible intended diagram for such a question structure.
– Resistors in parallel: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances ($1/R_{total} = 1/R_1 + 1/R_2 + \dots$). For two resistors, this simplifies to $R_{total} = (R_1 \times R_2) / (R_1 + R_2)$.
– Assuming common resistor values (like 3 and 5 ohm) and standard circuit configurations (like parallel) often helps in solving such MCQs when the diagram is missing but options are specific.
Series combination of resistors ($R_{total} = R_1 + R_2 + \dots$) results in a resistance greater than any individual resistance. Parallel combination results in a resistance smaller than the smallest individual resistance. In this case, 15/8 = 1.875 is smaller than both 3 and 5 ohm. Other standard configurations like series-parallel combinations could also result in fractional resistances, but the 3||5 parallel arrangement directly yields 15/8.