Question

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

Answer 1

Required Answer:

I)

Here, we are given resistances connected in series combination.

❒ The sum of individual resistances is equal to the total resistance in the series combination.

$\underline{\boxed{\sf{ {R}_{s} = {R}_{1} + {R}_{2} + {R}_{3} +. . . . {R}_{n} }}}$

• R₁ → 3 Ω
• R₂ → 3 Ω
• R₃ → 3 Ω

→ Rₛ = R₁ + R₂ + R₃

→ Rₛ = 3Ω + 3Ω + 3Ω

→ Rₛ = 9Ω

Hence, equivalent resistance between P and Q is 9Ω.

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II)

Here, we are given resistances connected in parallel combination.

❒The sum of reciprocals of all the individual resistances is equal to the reciprocal of the total resistance in the parallel combination.

$\underline{\boxed{\sf{ \dfrac{1}{{R}_{P}} = \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{2}} + \dfrac{1}{{R}_{3}} +. . . . \dfrac{1}{{R}_{n}} }}}$

• R₁ → 3 Ω
• R₂ → 3 Ω
• R₃ → 3 Ω

→ $\sf { \dfrac{1}{{R}_{P}}= \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}}$

→ $\sf { \dfrac{1}{{R}_{P}}= \dfrac{1+1+1}{3} }$ Ω

→ $\sf { \dfrac{1}{{R}_{P}}= \dfrac{3}{3} }$ Ω

→ $\sf { \dfrac{1}{{R}_{P}} = 1 }$ Ω

→ $\sf { {R}_{P} = \dfrac{1}{1} }$ Ω

→ $\sf { {R}_{P} = 1 }$ Ω

Hence, equivalent resistance between P and Q is 1Ω.

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Answer 2

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

$\sf{Here \: Resistances \: are ,}$

$\sf• \: R_{1} =3 Ω$

$\sf• \: R_{2} =3Ω$

$\sf• \: R_{3} =3Ω$

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

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

$\sf ⇒ R_{s} =(3 + 3 + 3)Ω$

$\sf \therefore R_{s} =9Ω$

$\sf \pink{ \boxed{Answer : 9 Ω.}}$

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

$\sf{Here \: Resistances \: are ,}$

$\sf• \: R_{1} =3 Ω$

$\sf• \: R_{2} =3 Ω$

$\sf• \: R_{3} =3 Ω$

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

$\bf \red {\bigstar {\: \frac{1}{R_{p}} = \frac{1}{R _{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}+...+\frac{1}{R_{n}} }}$

$\sf⇒ \frac{1}{ R_{p} } = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$

$\sf⇒ \frac{1}{ R_{p} } = \frac{1 + 1 + 1}{3}$

$\sf⇒ \frac{1}{ R_{p} } = \frac{3}{3}$

$\sf⇒ \frac{1}{ R_{p} } = 1$

$\sf \therefore R_{p} = 1Ω$

$\sf \pink{ \boxed{Answer : 1 Ω.}}$