Question

If five resistances each of value 0.2 ohm are connected in series, what be the resultant resistance ?

Answer 1

AnsWer :

1Ω.

To FinD :

what be the resultant resistance ?

SolutioN :

Let,

• The Resistance be.
• R1 = 0.2.
• R2 = 0.2.
• R3 = 0.2.
• R4 = 0.2.
• R5 = 0.2.

✒ Condition :

• Given resistance connect in series combination.

$\rule{90}2$

→ Rs = R1 + R2 + R3 + R4 + R5 ... Rn.

→ Rs = 0.2 + 0.2 + 0.2 + 0.2 + 0.2.

→ Rs = 1Ω.

$\rule{90}2$

✎ Diagram :

----------( R1 )--( R2 )--( R3 )--( R4 )--( R5 )----------

✡ Therefore, the resultant resistance is 1 Ω.

Answer 2

Solution :

We have five resistance each of value is 0.2 ohm are connected in series.'

$\bigstar$Firstly, we know that resistance in series :

When two or more resistance are joined end to end so that same current flows through each one of them in turn, they are said to be connected in series.

$\boxed{\sf{R_s=R_1+R_2+R_3+............R_n}}}$

A/q

Since, five resistance are connected in series, then total resistance of the circuit;

$\longrightarrow\sf{R=0.2+0.2+0.2+0.2+0.2}\\\\\longrightarrow\sf{R=0.2\times 5}\\\\\longrightarrow\bf{R=1\Omega}$