Question

Calculate equivalent resistance in the following between points P and Q.​

Answer 1

1st one is series circuit and the 2nd one is parallel circuit.

Answer is 1 ohm.

hope it helps you.

Answer 2

`\sf{(i)  \: We \:  are \:  given  \: a  \: figure \:  of  \: Series \:  Circuit. }`

`\sf{Here \:  Resistances  \: are ,}`

`\sf• \:  R_(1)  =3  Ω`

`\sf•  \: R_(2)  =3Ω`

`\sf•  \: R_(3)  =3Ω`

`\sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Series  \: Circuit, }`

`\bf \red {\bigstar  {\: R_(s)  =  R _(1)  +  R _(2)  +  R_(3)+...+ R_(n)}}`

`\sf ⇒ R_(s) =(3 + 3 + 3)Ω`

`\sf \therefore R_(s)  =9Ω`

`\sf  \pink{ \boxed{Answer : 9 Ω.}}`

`\\  \\`

`\sf{(ii)  \: We \:  are \:  given  \: a  \: figure \:  of  \: Parallel \:  Circuit. }`

`\sf{Here \:  Resistances  \: are ,}`

`\sf• \:  R_(1)  =3  Ω`

`\sf• \:  R_(2)  =3  Ω`

`\sf• \:  R_(3)  =3  Ω`

`\sf{We \:  know  \: the  \: formula  \: of  \: the  \: Equivalent \:  Resistance \:  for  \: Parallel  \: Circuit, }`

`\bf \red {\bigstar  {\:  (1)/(R_(p))  =   (1)/(R _(1))  +   (1)/(R_(2))  +  (1)/(R_(3))+...+(1)/(R_(n)) }}`

`\sf⇒ (1)/( R_(p) )  =  (1)/(3)  +  (1)/(3)  +  (1)/(3)`

`\sf⇒ (1)/( R_(p) )  =  (1 + 1 + 1)/(3)`

`\sf⇒ (1)/( R_(p) )  =  (3)/(3)`

`\sf⇒ (1)/( R_(p) )  =  1`

`\sf \therefore R_(p)   =  1Ω`

`\sf  \pink{ \boxed{Answer : 1 Ω.}}`