Question

three resistance 2 ohm , 4 ohm and 8 ohm are connected in parallel.if total current from battery is 8 amp then find total resistance , total potential of battery and current through each resistance​

Answer 1

Answer:

The total resistance will be 8/7

Answer 2

GIVEN :-

• Resistance, $\sf R_1 = 2 \ \Omega$

• Resistance, $\sf R_2 = 4 \ \Omega$

• Resistance, $\sf R_3 = 8 \ \Omega$

• Current, $\sf I = 8 \ A$

TO FIND :-

• The total potential of the battery.

• Current through each resistance.

SOLUTION :-

• Effective resistance when resistors are connected in parallel,

               $\bf \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

        $\longrightarrow\sf \dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} \\\\\longrightarrow \dfrac{1}{R} = \dfrac{4}{8}+\dfrac{2}{8}+\dfrac{1}{8} \\\\\longrightarrow\dfrac{1}{R}=\dfrac{7}{8} \\\\\longrightarrow R=\dfrac{8}{7} \\\\\longrightarrow \bf R = 1.14 \ \Omega$

 

  We know that ,

     V = IR

 Total potential of the battery , V = 8 × 1.14

                                                   = 9.12 V

• Current through the first resistor

         $\longrightarrow \sf I_1 = \dfrac{V}{R_1} \\\\\longrightarrow I_1 = \dfrac{9.12}{2} \\\\\longrightarrow\bf I_1 = 4.56 \ A$

• Current through the second resistor

         $\longrightarrow\sf I_2 = \dfrac{V}{R_2}\\\\\longrightarrow I_2 = \dfrac{9.12}{4}\\\\\longrightarrow \bf I_2 = 2.28 \ A$

• Current through the third resistor

         $\longrightarrow\sf I_3 = \dfrac{V}{R_3}\\\\\longrightarrow I_3 = \dfrac{9.12}{8}\\\\\longrightarrow\bf I_3 = 1.14 \ A$