Question

please answer this question ?

Class - 10th
Chapter - Electricity

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Answer 1

$\huge{\boxed {\mathbb {QUESTION}}}$

a student has two resistors of 2Ω and 3 Ω . he has to put one of them in place of R2 . show in the circuit The current that he needs in the entire circuit is exactly 7A. Show by calculation which of the two resistors she should choose

$\huge{\boxed {\mathbb {ANSWER}}}$

${\boxed {\mathbb {GIVEN}}}$

$Battery\: source (V) =12V$

$Total\: current (I) =7A$

$R_1=4Ω$

${\boxed {\mathbb {TO\:PROVE}}}$

$Which\:resistor \:choose \:in\:R_2$

${\boxed {\mathbb {SOLUTION}}}$

${\boxed {\boxed {V=IR}}}$

$V=Battery\: source$

$I=current$

$R=resistance$

$Substitute\: the\: values$

$\implies 12=7\times R$

$\implies R=\frac{12}{7}$

$R_1\:andR_2\:are\:in\:parallel$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$Substitute \frac{1}{R}=\frac{12}{7}$

$\frac{1}{\frac{12}{7}}=\frac{1}{4}+\frac{1}{R_2}$

$\implies\frac{1}{12}\times 7=\frac{1}{4}+\frac{1}{R_2}$

$\implies\frac{7}{12}=\frac{1}{4}+\frac{1}{R_2}$

$\implies \frac{7}{12}-\frac{1}{4}=\frac{1}{R_2}$

$\implies \frac{7-3}{12}=\frac{1}{R_2}$

$\implies \frac{4}{12}=\frac{1}{R_2}$

$\implies \frac{1}{3}=\frac{1}{R_2}$

$\implies{\boxed {\boxed {R_2=3Ω}}}$

$\therefore 3Ω\: resistor\: should\: be \:placed\: in\: R_2$

$\huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU }}}$

_________________________________________

$\huge {\boxed {\mathbb {EXTRA \:INFORMATION}}}$

$Formulas$

$I=\frac{V} {R}$

$V=\frac{W} {Q}$

$V=IR$

${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}$

Answer 2

V=IR

12=7R

R=12/7

Resistances are parallel, so

1/R1 + 1/R2 = 7/12

1/4 + 1/R2= 7/12

1/R2= 7/12-1/4

1/R2= 4/12

1/R2=1/3

R2=3ohm

Hence, use 3 ohm resistance.

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