Question

. Two resistors of resistance 4Ω and 8Ω are connected in series and parall, find the

equivalent resistances in both cases?​

Answer 1

Explanation:

In series connection R

n

=R

1

+R

2

+R

3

+R

4

=5+6+4+8=23Ω

In the parallel connection

R

n

1

=

5

1

+

6

1

+

4

1

+

8

1

=0.74Ω

−1

R

n

=1.34Ω

Answer 2

Solution:-

Case 1 :-

$\rm \to \: if \: two \: \: resistor \: \: R_1 \: and \: R_2 \: are \: connect \: in \: series$

Formula of series combination

$\rm \to \: R_e = R_1 + R_2$

so we have

$\rm \to \: R_1 = 4 \: ohm$

$\rm \to \: R_2 = 8 \: ohm$

Put the value on formula

$\rm \to \: R_e =( 4 + 8)ohm$

$\rm \to \: R_e \: = 12 \: ohm$

Case:- 2

$\rm \to \: if \: two \: \: resistor \: \: R_1 \: and \: R_2 \: are \: connect \: in \: parallel$

Formula for parallel combinations

$\rm \to \dfrac{1}{ R_e} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\rm \: \dfrac{1}{R_e} = \dfrac{1}{4} + \dfrac{1}{8}$

Taking lcm

$\rm \to \dfrac{1}{R_e} = \dfrac{1 \times 2}{4 \times 2} + \dfrac{1}{8}$

$\rm \to \: \dfrac{1}{R_e} = \dfrac{2 + 1}{8}$

$\rm \to \: \dfrac{1}{R_e} = \dfrac{3}{8}$

answer is

$\rm \to \: R_e = \dfrac{8}{3}$