Home
Class 11
PHYSICS
The values of two resistors are R(1)=(6+...

The values of two resistors are `R_(1)=(6+-0.3)kOmega` and `R_(2)=(10+-0.2)kOmega`. The percentage error in the equivalent resistance when they are connected in parallel is

A

`5.125%`

B

`2%`

C

`10.125%`

D

`7%`

Text Solution

Verified by Experts

The correct Answer is:
C
`R=(R_1R_2)/((R_1+R_2))=(10xx6)/(10+6)=60/10kOmega=3.75 KOmega`
`implies (DeltaR)/R=(DeltaR_1)/R_2+(DeltaR_2)/R_2+(Delta(R_1+R_2))/((R_1+R_2))`
`=(0.3)/(6)+(0.2)/(10)+((0.3 +0.2))/(10+6) = 0.10125`
`implies (DeltaR)/R = 10.125%`
Promotional Banner

Similar Questions

Explore conceptually related problems

Two resistor R_(1) = (24 +- 0.5)Omega and R_(2) = (8 +- 0.3)Omega are joined in series, The equivalent resistence 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 resistance R_(1)=100pm 3Omega and R_(2)=200pm4Omega are connected in seriesg. What is their equivalent resistance?

Two resistors of resistances R_(1)=(300+-3) Omega and R_(2)=(500+-4) Omega are connected in series. The equivalent resistance of the series combination is

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.

Two resistors R_1 = (4 pm 0.8) Omega and R_2 (4 pm 0.4) Omega are connected in parallel. The equivalent resistance of their parallel combination will be :

Given Resistance R_(1) = (8 +- 0.4) Omega and Resistence, R_(2) = (8 +- 0.6) Omega What is the net resistence when R_(1) and R_(2) are connected in series?

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 carbon resistances are given by R_(1)=(4pm0.4) ohm and R_(2)=(12pm0.6)Omega . What is their net resistance with percentage error, if they are connected in series?