Question

Three resistor 1ohm, 2ohm, 3ohm, are combined in series what is the total resistance of the combination

Answer 1

Given:

Three resistors 1Ω, 2Ω and 3Ω are connected in series in a circuit

To Find:

Total resistance of the combination

Solution:

We know that,

• In a series combination of n resistances, equivalent or total resistance R is equal to the sum of of individual resistances

$\pink{\boxed{\bf{R=R_{1}+R_{2}+R_{3}+....+R_{n}}}}$

$\rule{190}{1}$

Let the total resistance of given combination be 'R'

So,

$\longrightarrow\rm{R=1+2+3}$

$\longrightarrow\rm\green{R=6\:\Omega}$

Hence, the total resistance of given combination is 6 Ω.

$\rule{190}{1}$

Additional Information:-

In a series combination of resistances,

• Current through each resistance is same
• Potential difference across each resistance is different and the main potential is equal to the sum of individual potential differences across every resistance

Answer 2

GIVEN :-

Three resistors of resistance 1Ω, 2Ω and 3Ω are connected in series.

TO FIND :-

The total resistance of the combination or the equivalent resistance of the combination.

FORMULA TO BE USED :-

For resistors connected in series,

$\boxed{\sf{R_{Equivalent}=R_1+R_2+R_3+R_4+.....+R_n}}$

SOLUTION :-

So now, for the given question,

$\sf{R_{Equivalent}=R_1+R_2+R_3}$

$\sf{\implies R_{Equivalent}=1+2+3}$

$\boxed{\boxed{\sf{\implies R_{Equivalent}=6\Omega}}}$

$\underline{\rule{240}{2}}$

*Extra info :-

For resistors connected in parallel,

$\sf{\dfrac{1}{R_{Equivalent}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.....+\dfrac{1}{R_n} }$

• The distribution of voltage in this connection is not uniform.