Question

A parallel combination of three resistors take a current of5A from 20v supply.if the two resistor are 10ohm &8ohm ,find the value of third once?

Answer 1

$\huge \mathfrak{solution}$

Given:

• A parallel combination of three resistors .In which two. are given : 10 Ω and 8 Ω.

• Current = 5A

• V = 20V

To find :

• Third resistor .

Answer:

• First of all find the Total (equivalent) resistance:

Let total resistance be R

As we know that:

$\star \boxed{v = {i}{r} }$

=> R= v/i

=>R= 20/5= 4 Ω

Now let the third resistance be r3

As we know that , In parallel combination we use :

$\star \: \boxed{ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}.....}$

Here :

• r1= 10 Ω

•r2=8 Ω

• r(eq)= 4 Ω

$\implies \: \frac{1}{4} = \frac{1}{10} + \frac{1}{8} + \frac{1}{r_3} \\ \\ \implies \: \frac{1}{r_3} = \frac{1}{4} - \frac{1}{10} - \frac{1}{8} \\ \\ \implies \: \frac{1}{r_3} = \frac{10 - 4 - 5}{40} \\ \\ \implies \: \frac{1}{r_3} = \frac{1}{40} \\ \\ \implies \: r_3 = 40 \Omega$

Hence , the value of third resistor is 40 Ω

Answer 2

$</p><p>\huge{\tt{\pink{Hello!!! }}}$

$\huge{\fbox{\fbox{\rightarrow {\mathfrak {\green {Question \: - }}}}}}$

• A parallel combination of three resistors take a current of5A from 20v supply.

• If the two resistor are 10ohm &8ohm ,find the value of third one.

$\huge{\fbox{\fbox{\rightarrow {\mathfrak {\green {Answer \: - }}}}}}$

$</p><p>\huge{\boxed{\tt{\purple{Given \: - }}}}$

$\tt{ \red{Type \: of \: Combination \: - \: Parallel }} \\ \\ \tt{ \green{rR1 = 10 \: ohm}} \: \\ \\ \tt{ \pink{R2 = 8 \: ohm}} \\ \\ \tt{ \blue{\: I \: = \: 5 \: A }} \: \\ \\ \tt{ \green{ V\: = \: 20 \: V\: }} \\ \\ \tt{ \purple{ \boxed{ \boxed{R \: = \: \frac{I}{V} \: = \: 4 \: ohm. }}}}$

Now let the third resistance be R3

As we know that , In parallel combination we use :

$</p><p>\tt{\blue{\huge{\star \: \boxed{ \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}.....} }}}}$

Here :

• R1 = 10 Ω

• R2 =8 Ω

• R(eq)= 4 Ω

$</p><p>\tt{\blue{\huge{ \implies \: \frac{1}{r_3} = \frac{1}{4} - \frac{1}{10} - \frac{1}{8} }}}$

$</p><p>\tt{\purple{\huge{ \implies \: r_3 = 40 \Omega}}}$