Home
Class 12
PHYSICS
Two resistance R(1) and R(2) are connect...

Two resistance `R_(1)` and `R_(2)` are connected in (i) series and (ii) parallel. What is the equivalent resistance with limit of possible percentage error in each case of `R_(1)=5.0+-0.2Omega` and `R_(2)=10.0+-0.1Omega`

A

`15 Omega pm 2% , 3.3 Omega pm 3%`

B

`25 Omega pm 25% , 3.3 Omega pm 2%`

C

`15 Omega 3% , 3.3 Omega pm 2%`

D

`3.3 Omega pm 2 % , 15 Omega pm 3%`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • ERRORS , THEORY

    MOTION|Exercise EXERCISE 2 (LEVEL -I) (SECTION A)|8 Videos
  • ERRORS , THEORY

    MOTION|Exercise EXERCISE 2 (LEVEL -I) (SECTION B)|7 Videos
  • ERRORS , THEORY

    MOTION|Exercise EXERCISE 1 (SECTION B)|7 Videos
  • ELECTROSTATICS-II

    MOTION|Exercise Exercise - 4 (Level - II)|30 Videos
  • FLUID

    MOTION|Exercise Exercise-4 level-II|5 Videos

Similar Questions

Explore conceptually related problems

Two resistance R_(1)=100pm 3Omega and R_(2)=200pm4Omega are connected in seriesg. What is their equivalent resistance?

Two conductors of resistance R_(1) (50 +- 2) ohm and R_(2) = (100 +- 4) ohm are connected in (a) series and (b) parallel. Find the equivalent resistance.

Tow resistance R_(1) = 50 pm 2 ohm and R_(2) = 60 pm connected in series , the equivalent resistance of the series combination in

If resistors of resistance R_(1) and R_(2) are connected in parallel,then resultant resistance is "

Two resistor of resistance R_(1)=(6+-0.09)Omega and R_(2)=(3+-0.09)Omega are connected in parallel the equivalent resistance R with error (in Omega)

Two resistances R_(1) = 100 +- 3 Omega and R_(2) = 200 +- 4 Omega are connected in series . Find the equivalent resistance of the series combination.

108 . If two resistors of resistances R_(1)= (4 +- 0.5 )Omega and R_(2) =(16+-0.5)Omega are connected (i) in series and (ii) in parallel, find the equivalent resistance in each case with limits of percentage error.

Two resistors of resistance R_(1) and R_(2) having R_(1) gt R_(2) are connected in parallel. For equivalent resistance R, the correct statement is

Two resistances R_1 = (16 +- 0.3) ohm and R_2 = (48 +- 0.5) ohm are connected in parallel. Find the total resistance of the combination and maximum percentage error.