Two resistances R₁ = 100 ± 3Ω and R₂= 200 ± 4Ω are connected in series. What is their equivalent resistance ?

The correct option is (300 ± 7)Ω.

Step-by-step Explanation:

When two resistors are connected in series, their equivalent resistance is the sum of their individual resistances. The formula is expressed as:

$R_s=R_1+R_2$

Given values in the question:
$R_1=100\pm 3\Omega$
$R_2=200\pm 4\Omega$

Here, the nominal (measured) values of the resistances are:
$R_1=100\Omega$
$R_2=200\Omega$

The absolute errors associated with each resistor are:
$\Delta R_1=3\Omega$
$\Delta R_2=4\Omega$

1. Find the Nominal Equivalent Resistance:
Add the nominal values together:
$R_s=100+200=300\Omega$

2. Find the Total Error in the Series Combination:
When adding physical quantities, their absolute errors must be added together to find the maximum possible uncertainty:
$\Delta R_s=\Delta R_1+\Delta R_2$

Substituting the errors:
$\Delta R_s=3+4=7\Omega$

3. Combine the Nominal Value and Total Error:
Writing the equivalent resistance with its uncertainty gives:
$R_s=(300\pm 7)\Omega$