Question

please answer this question ?

Class - 10th
Chapter - electricity

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Answer 1

The overall current needed = 9A.

The voltage is 12V

Hence by Ohm’s Law V=IR

The resistance for the entire circuit = 12/9Ω = 4/3Ω = R

R1 and R2 are in parallel. Hence, R= $\frac{(r1r2)}{r1 + r2}$

After putting their values we get r2 = 2Ω

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Answer 2

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

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

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

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

• $Battery\: source (V) =12V$
• $Total\: current (I) =7A$
• $R_1=4Ω$

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

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

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

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

• $V=Battery\: source$
• $I=current$
• $R=resistance$

$Substitute\: the\: values$

$\implies 12=7\times R$

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

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

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

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

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

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

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

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

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

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

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

$\implies{\boxed {\boxed {R_2=3Ω}}}$

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

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

_________________________________________

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

$Formulas$

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

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

$V=IR$

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