Question

Four identical resistors each of R=4 are joined in a circuit......Find the rating of ammeter if cell B has e.m.f of 2 volt
The Answer is 3/8 A
I want the explanation of this came with correct redrawn diagram otherwise i will have to report
Please tell the answer

Answer 1

Answer:

$\bigstar{\bold{Current=\dfrac{3}{8}\:A}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• The resistors are identical and is 4 Ω each.
• EMF of the cell is 2 V

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Current (I) in the circuit

$\Large{\underline{\underline{\bf{Figure:}}}}$

The redrawn diagram is given in the attachment

$\Large{\underline{\underline{\bf{Solution:}}}}$

➜ Find the total resistance through the 3 resistors connected in parallel

    $\dfrac{1}{R_p}=\dfrac{1}{R_1} +\dfrac{1}{R_2}+\dfrac{1}{R_3}$

    $\dfrac{1}{R_p}=\dfrac{1}{4} +\dfrac{1}{4} +\dfrac{1}{4} =\dfrac{3}{4}$

    $R_p=\dfrac{4}{3} \Omega$

➜ Next find the total resistance between the parallel connection and series connection

    $R=R_p+R_4$

    $R=\dfrac{4}{3}+4$

   $R=\dfrac{4+12}{3}$

   $R=\dfrac{16}{3} \Omega$

➜ By Ohm's law current in a circuit is given by

    $I=\dfrac{V}{R}$

➜ Substituting the datas we get,

    $I=\dfrac{2}{16/3} =\dfrac{6}{16} =\dfrac{3}{8}A$

$\boxed{\bold{Current=\dfrac{3}{8}\:A}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

➜ For resistors connected in  series total resistance is given by

    $R=R_1+R_2+R_3+........+R_n$

➜ For resistors connected in parallel, total resistance is given by

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