Question

the equivalent resistance of circuit diagram in 2 ohm. calculate resistance of r1 ​

Answer 1

Question :

The equivalent resistance of circuit diagram is 2Ω. Calculate resistance of $\sf\:r_1$

$\rule{200}2$

Before solving the question , Let's know about some Combination of resistors .

• Resistors in series and parallel :

1) Resistors in series :

For n resistors connected in series

the equivalent Resistance $R_{eq}$ is given by

$R_{eq}=R_{1}+R_{2}+R_{3}+...+R_{n}$

2) Resistors in parallel:

For n resistors connected in parallel

the equivalent Resistance $R_{eq}$ is given by

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1} } + \dfrac{1}{R_{2} } + \dfrac{1}{R_{3} } + ...... \dfrac{1}{R_{n} }$

Solution :

We have,

$\sf\:r_2=6$Ω

$\sf\:r_3=12$Ω

$\sf\:r_1=?(To\:Find)$

and $\sf\:r_{eq}=2$

Clearly, $\sf\:r_1,r_2\:and\:r_3$all are connected in parallel combination .

Then ,

$\sf\dfrac{1}{r_{eq}}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}$

$\sf\implies\dfrac{1}{2}=\dfrac{1}{r_1}+\dfrac{1}{6}+\dfrac{1}{12}$

$\sf\implies\dfrac{1}{2}=\dfrac{12+2r_1+r_1}{12r_1}$

$\sf\implies\dfrac{1}{2}=\dfrac{12+3r_1}{12r_1}$

$\sf\implies\:12r_1=24+6r_1$

$\sf\implies\:6r_1=24$

$\sf\implies\:r_1=4$Ω

Therefore, The vale of $\sf\:r_1=4$Ω

Answer 2

Answer:

Question :

The equivalent resistance of circuit diagram is 2Ω. Calculate resistance of \sf\:r_1r

1

\rule{200}2

Before solving the question , Let's know about some Combination of resistors .

• Resistors in series and parallel :

1) Resistors in series :

For n resistors connected in series

the equivalent Resistance R_{eq}R

eq

is given by

R_{eq}=R_{1}+R_{2}+R_{3}+...+R_{n}R

eq

=R

1

+R

2

+R

3

+...+R

n

2) Resistors in parallel:

For n resistors connected in parallel

the equivalent Resistance R_{eq}R

eq

is given by

\dfrac{1}{R_{eq}} = \dfrac{1}{R_{1} } + \dfrac{1}{R_{2} } + \dfrac{1}{R_{3} } + ...... \dfrac{1}{R_{n} }

R

eq

1

=

R

1

1

+

R

2

1

+

R

3

1

+......

R

n

1

Solution :

We have,

\sf\:r_2=6r

2

=6 Ω

\sf\:r_3=12r

3

=12 Ω

\sf\:r_1=?(To\:Find)r

1

=?(ToFind)

and \sf\:r_{eq}=2r

eq

=2

Clearly, \sf\:r_1,r_2\:and\:r_3r

1

,r

2

andr

3

all are connected in parallel combination .

Then ,

\sf\dfrac{1}{r_{eq}}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}

r

eq

1

=

r

1

1

+

r

2

1

+

r

3

1

\sf\implies\dfrac{1}{2}=\dfrac{1}{r_1}+\dfrac{1}{6}+\dfrac{1}{12}⟹

2

1

=

r

1

1

+

6

1

+

12

1

\sf\implies\dfrac{1}{2}=\dfrac{12+2r_1+r_1}{12r_1}⟹

2

1

=

12r

1

12+2r

1

+r

1

\sf\implies\dfrac{1}{2}=\dfrac{12+3r_1}{12r_1}⟹

2

1

=

12r

1

12+3r

1

\sf\implies\:12r_1=24+6r_1⟹12r

1

=24+6r

1

\sf\implies\:6r_1=24⟹6r

1

=24

\sf\implies\:r_1=4⟹r

1

=4 Ω

Therefore, The vale of \sf\:r_1=4r

1

=4 Ω