Two equal resistors R are connected in parallel, and a battery of 12 V

Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination. A dc current of 100 mA flows through the circuit as shown below :
The value of R is

[amp_mcq option1=”120 Ω” option2=”240 Ω” option3=”60 Ω” option4=”100 Ω” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

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Two equal resistors R are connected in parallel. The equivalent resistance (Req) of two equal resistors in parallel is given by Req = R/2. According to Ohm’s Law, the voltage (V) across the circuit is equal to the current (I) flowing through it multiplied by the equivalent resistance (Req): V = I * Req.
Given V = 12 V and I = 100 mA = 0.1 A.
So, 12 V = 0.1 A * (R/2).
Rearranging the equation to solve for R:
12 = 0.1 * R / 2
12 = 0.05 * R
R = 12 / 0.05
R = 12 / (5/100)
R = 12 * (100/5)
R = 12 * 20
R = 240 Ω.

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: 1/Req = 1/R1 + 1/R2 + … For two equal resistors R, 1/Req = 1/R + 1/R = 2/R, so Req = R/2.

Ohm’s Law (V = IR) is fundamental to analyzing simple electric circuits. Units must be consistent (Volts, Amperes, Ohms). Milliamperes (mA) must be converted to Amperes (A) by dividing by 1000 (100 mA = 0.1 A).

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