Question

find the effective resistance when two resistance 6 ohm and 2 ohm are connected in series and parallel​

Answer 1

Answer:

Series= 8 ohm

parallel= 1.5 ohm

Explanation:

In series connection

R = R1 + R2

R = 6+2 = 8 ohm

In parallel combination

1/R = 1/R1 + 1/R2

1/R = 1/6+1/2

1/R = 1 + 3 / 6

1/R = 4/6

R = 6/4 = 3/2= 1.5 ohm

Answer 2

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• Two resistance of 6Ω and 3Ω are connected in series and parallel .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The effective resistance in both cases.

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

We know that when resistors are connected in series then the net resistance is equal to the sum of individual resistances . Whereas when the Resistances are connected in parallel then the reciprocal of net resistance is equal to the the sum of reciprocal of individual resistances.

➥ Case 1 : In series :-

$\sf\dashrightarrow R_{(net)}= R_1+ R_2 \\\\\\\sf\dashrightarrow R_{(net)}= ( 6 + 2 ) \Omega \\\\\\\sf\dashrightarrow \boxed{\orange{\sf R_{(net)}= 8 \Omega }}$

$\rule{200}4$

➥ Case 2 : In parallel :-

$\sf\dashrightarrow \dfrac{1}{R_{(net)}}= \dfrac{1}{R_1}+ \dfrac{1}{R_2 } \\\\\\\sf\dashrightarrow R_{(net)}= \dfrac{R_1R_2}{R_1+R_2} \\\\\\\sf\dashrightarrow R_{(net)}= \dfrac{ (2*6)\Omega^2}{(2+6)\Omega } \\\\\\\sf\dashrightarrow R_{(net)}= \dfrac{ 12}{8}\Omega \\\\\\\sf\dashrightarrow \boxed{\orange{\sf R_{(net)}= 1 .5 \Omega }}$

$\rule{200}4$