Two equal resistors R are connected in parallel, and a battery of 12 V is connected across this combination. A d.c. current of 100 mA flows through the circuit as shown below. The value of R is
- (a)120 Ω
- (b)240 Ω
- (c)60 Ω
- (d)100 Ω
Correct — B, 240 Ω. The 100 mA is the total current drawn from the battery, so the combination's resistance is R_total = V / I = 12 V / 0.1 A = 120 Ω. Two equal resistors in parallel give R_total = R/2, so R/2 = 120 Ω, which means each R = 240 Ω.
- (a)120 Ω — 120 Ω is the resistance of the whole parallel combination (V/I), not the value of one resistor — you must still double it to get R.
- (c)60 Ω — 60 Ω would result from halving 120 Ω instead of doubling it — the wrong direction for a parallel pair.
- (d)100 Ω — 100 Ω simply misreads the 100 mA current figure as if it were the resistance; the two are different quantities.
For resistors in parallel, the equivalent resistance is smaller than either resistor: 1/R_total = 1/R1 + 1/R2. For two equal resistors this simplifies to R_total = R/2. Combined with Ohm's law (V = I R), the battery current fixes the combination's resistance, from which the individual resistor value follows.
The trap is to stop at V/I = 120 Ω and pick that as R. But 120 Ω is the parallel combination; each equal resistor must be twice that, so R = 240 Ω.
- Ohm's law: V = I x R, so R_total = V/I = 12/0.1 = 120 Ω.
- Two equal resistors in parallel: R_total = R/2.
- Therefore R = 2 x R_total = 2 x 120 = 240 Ω.
- Parallel resistance is always less than the smallest branch resistor.
The 100 mA is the total current, so R_total = 120 Ω and each resistor R = 240 Ω.
- Treating V/I as the individual resistor rather than the parallel combination.
- Forgetting to convert 100 mA to 0.1 A before applying Ohm's law.
Asked as a numerical Ohm's-law-plus-combination problem; identify whether the given current is total or branch current.
Two wires have their lengths, diameters and resistivities, all in the ratio of 1 : 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is
- (a) 10 ohms
- (b) 5 ohms
- (c) 20 ohms
- (d) 40 ohms
Answer(a) 10 ohms
Same topic — resistance and Ohm's law. UPSC 2001 reasons about resistance from wire dimensions and resistivity; this NDA item solves for the resistor value in a parallel circuit.
If three resistors of 1 Ohm each connect in parallel to each other the resultant resistance is
- (a) 1 Ohm
- (b) 1⁄3 Ohm
- (c) 3 Ohm
- (d) 9 Ohm
Answer(b) 1⁄3 Ohm
Same concept — equal resistors in parallel. That NDA item gives the resultant of three equal resistors (R/3); this one works backward from the combination to each resistor (R = 2 x R_total).
- practice — not a real PYQ
Two equal resistors of 240 Ω each are connected in parallel. Their combined resistance is
- (a)480 Ω
- (b)240 Ω
- (c)120 Ω
- (d)60 Ω
Answer(c) 120 Ω — for two equal resistors in parallel, R_total = R/2.
- practice — not a real PYQ
A 12 V battery drives a total current of 100 mA through a resistive load. The load's resistance is
- (a)1.2 Ω
- (b)12 Ω
- (c)120 Ω
- (d)1200 Ω
Answer(c) 120 Ω — R = V/I = 12 / 0.1 = 120 Ω.