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Answer:
Answer:
Given :-
The resístance of each resístors is 10 ohm resistors.
To Find :-
What is the maximum resístance make available using two resístors.
Formula Used :-
\clubsuit♣ Equívalent Resístance for series connection :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{R_{eq} =\: R_1 + R_2\: +\: . . . .\: +\: R_n}}}\\\end{gathered}↦Req=R1+R2+....+Rn
where,
\sf R_{eq}Req = Equívalent Resístance
\sf R_1R1 = Resístance of resístors R₁
\sf R_2R2 = Resístance of resístors R₂
Solution :-
Given :
\begin{gathered}\bigstar\: \rm{\bold{Resistance\: of\: resistors\: (R_1) =\: 10\: \text{\O}mega}}\\\end{gathered}★Resistanceofresistors(R1)=10Ømega
\bigstar\: \rm{\bold{Resistance\: of\: resistors\: (R_2) =\: 10\: \text{\O}mega}}★Resistanceofresistors(R2)=10Ømega
According to the question by using the formula we get,
\longrightarrow \sf R_{eq} =\: R_1 + R_2⟶Req=R1+R2
\longrightarrow \sf R_{eq} =\: 10\: \text{\O}mega + 10\: \text{\O}mega⟶Req=10Ømega+10Ømega
\longrightarrow \sf R_{eq} =\: (10 + 10)\: \text{\O}mega⟶Req=(10+10)Ømega
\longrightarrow \sf R_{eq} =\: (20)\: \text{\O}mega⟶Req=(20)Ømega
\longrightarrow \sf\bold{\red{R_{eq} =\: 20\: \text{\O}mega}}⟶Req=20Ømega
\therefore∴ The maximum resístance is 20 Ω . It is connected by equívalent resístance of series connection.
\begin{gathered}\\\end{gathered}
EXTRA INFORMATION :-
\clubsuit♣ Equívalent Resístance for párallel connection :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{R_{eq}} =\: \dfrac{1}{R_1} + \dfrac{1}{R_2}\: +\: . . . .\: +\: \dfrac{1}{R_n}}}}\\\end{gathered}↦Req1=R11+R21+....+Rn1
where,
\sf R_{eq}Req = Equívalent Resístance
\sf R_1R1 = Resístance of resístors R₁
\sf R_2R2 = Resístance of resístors R₂
Answer:
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