Question

- Derivation of
series combination of resistors​

Answer 1

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$\small{\underline{\green{\sf{Solution :}}}}$

As we know that in series combination

$\star {\boxed{\sf{V_s \: = \: V_1 \: + \: V_2 \: + \: ......... \: + \: V_n}}}$

Where,

• V is potential difference
• And Vn is Potential Difference til n terms

As we also know that V = IR,

Put values of V. Where as in series I (current) is equal for all the resistance.

So,

$\implies {\sf{IR_s \: = \: IR_1 \: + \: IR_2 \: + \: ........ \: + \: IR_n}} \\ \\ \implies {\sf{\cancel{I}(R_s) \: = \: \cancel{I}(R_1 \: + \: R_2 \: + \: ........ \: R_n)}} \\ \\ \implies {\sf{R_s \: = \: R_1 \: + \: R_2 \: + \: ......... \:R_n}}$

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Additional Information :

• In series combination Current at all the resistors is provided equally .

• Whereas Potential Difference varies from resistors to resistors.

• Series combination provides the highest over all resistances.

Answer 2

Answer:

To derive:

Equivalent Resistance in series combination of resistors

Derivation:

Let's consider n number of resistors attached in series combination across

a battery of voltage V.

Let the resistors be R1 , R2, R3 ,.....n resistances

We know that the total Voltage will be equal to the sum of individual voltage drops along each Resistances

$V = \: V1 + V2 + ......Vn$

$= > V = (i \times R1) + (i \times R2) \times ...(i \times Rn)$

$= > V = i \times (R1 + R2 + ...Rn)$

$= > i \times R \: eq. = i \times (R1 +R2 ....)$

Cancelling i on both sides:

$= > R \: eq. = R1 + R2 + ...Rn$

So , we can represent the Equation as:

$\boxed{R \: eq. = \Sigma(Ri)}$