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  • How can three resistors of resistances 2 omega, 3 omega, and 6 omega be connected to give a total resistance of (a) 4 omega (b) 1 omega ?

   How can three resistors of resistances 2\Omega ,3\Omega , \; and \; 6\Omega  be connected to give a total resistance of

  (a)4\Omega , (b)1\Omega ?

Answers (1)

best_answer

(a) 4 Ohm

R_1=3,R_2=6,R_3=2

    R=Equivalent  resistance

    \frac{1}{R_1_2}=\frac{1}{R_1}+\frac{1}{R_2} 

\Rightarrow \frac{1}{R_1_2}=\frac{1}{3}+\frac{1}{6}

\Rightarrow \frac{1}{R_1_2}=\frac{2+1}{6}

\Rightarrow R_1_2=\frac{6}{3}=2

R=R_1_2+R_3=2+2=4

(b).1 Ohm

  R=Equivalent  resistance

Connect all three resistors in parallel

    \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} 

\Rightarrow \frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}

\Rightarrow \frac{1}{R}=\frac{6+3+1}{6}

\Rightarrow R=\frac{16}{6}=1\Omega

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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