Question

SOLVE THIS QUESTION FRIENDS , SUBJECT - ELECTRIC CURRENT 10th​

Answer 1

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Q1)The three conductors having resistance values 2ohm, 3 and 4 oh are connected in parallel :

$\: \: \: \: \boxed{ \bf \pink{to \: find \to\: equivalent \: resistance}}$

$\huge{\mathtt{\purple{A}\green{N}\pink{S}\blue{W}\purple{E}\green{R}\pink{!}\blue{!}}}$

if the conductor are connected in series then,

$\implies\boxed{\bf\red{R_1=2 \: R_2=3 \: and R_3=4 ohm}}$

$\mapsto\boxed{\bf\green{Equivalent \: resistance(R_net )\implies \: R_1+R_2+R_3}}$

$\: \: \mapsto \ \mathfrak{ \: by \: using \: this \: formula}$

$\to\displaystyle\bf{(R_net \: in \: series) =R_1+R_2+R_3}$

$\: \: \: \implies \boxed{ \rm{substitue \: the \: given \: values}}$

$\to\displaystyle\bf{R_net=2ohm+3ohm+4ohm}$

$\to\displaystyle\bf{R_net=(2+3+4)ohm}$

$\to\displaystyle\bf{R_net=9ohm}$

$\: \: \red \bigstar \boxed { \bf \blue{ \: equivalent \: resistance = 9ohm}}$

✯If the conductors are connected in parallel then ,

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{1}{R_1} +\dfrac{1}{R_2}+\dfrac{1}{R_3}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{3+2}{6}+\dfrac{1}{4}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{5}{6}+\dfrac{1}{4}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{20+6}{24}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{26}{24}}$

$\mapsto\bf{\dfrac{1}{R_net}=\cancel\dfrac{26}{24}}$

$\mapsto\bf{\dfrac{1}{R_net}=\dfrac{13}{12}}$

$\red\bigstar\boxed{\bf\green{Equivalent\: Resistance \: in \: parallel \: is\: \dfrac{13}{12}}}$

Answer 2

Question :

The three conductors having resistance 2 Ohm , 3 Ohm and 4 Ohm are connected in parallel . Find eq. resistance ?

Solution :

If resistors of resistances are connected in parallel ,

$\pink{\bigstar}\ \; \sf \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...$

Here ,

R₁ = 2 Ω , R₂ = 3 Ω , R₃ = 4 Ω

$\to \sf \dfrac{1}{R_{eq}}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\\\\\to \sf \dfrac{1}{R_{eq}}=\dfrac{3}{4}+\dfrac{1}{3}\ \qquad \\\\\to \sf \dfrac{1}{R_{eq}}=\dfrac{9+4}{12}\ \; \qquad \\\\\to \sf \dfrac{1}{R_{eq}}=\dfrac{13}{12}\ \; \qquad \; \\\\\to \sf R_{eq}=\dfrac{12}{13}\ \; \qquad \; \; \\\\\to \sf \pink{R_{eq}\cong 0.923\ \Omega}\ \bigstar \quad$

More Info :

If resistors of resistances are connected in series ,

$\sf \orange{\bigstar}\ \; R_{eq}=R_1+R_2+R_3+R_4+...$

If n equal resistors are connected in series ,

$\green{\bigstar}\ \; \sf R_{eq}=nR$