Question

please solve it A.S.A.P​

Answer 1

Explanation:

Resistance in series circuit : \begin{lgathered}R_s = R_1 + R_2 + R_3 + ...+ R_n\\\end{lgathered}

R

s

=R

1

+R

2

+R

3

+...+R

n

(equation 1)

Resistance in parallel circuit : \begin{lgathered}\frac{1}{R_p} = \frac{!}{R_1} + \frac{1}{R_2} +\frac{1}{R_3} +....+ \frac{1}{R_n}\\ \\\end{lgathered}

R

p

1

=

R

1

!

+

R

2

1

+

R

3

1

+....+

R

n

1

(equation 2)

Resistance in parallel circuit for two resistnces : R_p = \frac{R_1\times R_2}{R_1 + R_2}R

p

=

R

1

+R

2

R

1

×R

2

(equation 3)

Circuit CDE : Two series resistances of resistance R

So resistance in CDE = R + R = 2R

CE and CDE are in parallel of resistance 2R each.

So Resistance in Triangle CDE = R

So this CDE triangle becomes a resistor of resistance = R

Similarly for triangle BCE

BC and CE are in series with resistance R each

So BCE = 2R

Then

BCE and BE of equal resistance 2R in parallel make this triangle's resistance = R

Similarly for triangle ABE

AB and BE are in series with resistance R each

So ABE = 2R

Then

ABE and AE of equal resistance 2R in parallel make this triangle's resistance = R

\textbf{\Large So the whole circuit has total effective resistance = R}So the whole circuit has total effective resistance = R

Answer 2

Answer's in the pic............