3. In the given circuit, the effective resistancebetween points A and

1.	3. In the given circuit, the effective resistancebetween points A and E will be
Answer» The whole circuit has total effective RESISTANCE = R Step by step explanation: Resistance in SERIES circuit : $R_s = R_1 + R_2 + R_3 + ...+ R_n\\$ (equation 1) Resistance in PARALLEL circuit : $\frac{1}{R_p} = \frac{!}{R_1} + \frac{1}{R_2} +\frac{1}{R_3} +....+ \frac{1}{R_n}\\ \\$ (equation 2) Resistance in parallel circuit for two resistnces : $R_p = \frac{R_1\times R_2}{R_1 + R_2}$ (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 = $R_C_D_E = \frac{2R\times 2R}{2R + 2R} =$ 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 $\textbf{\Large So the whole circuit has total effective resistance = R}$

3. In the given circuit, the effective resistancebetween points A and E will be

Answer»

Step by step explanation:

Resistance in SERIES circuit : $R_s = R_1 + R_2 + R_3 + ...+ R_n\\$ (equation 1)

Resistance in PARALLEL circuit : $\frac{1}{R_p} = \frac{!}{R_1} + \frac{1}{R_2} +\frac{1}{R_3} +....+ \frac{1}{R_n}\\ \\$ (equation 2)

Resistance in parallel circuit for two resistnces : $R_p = \frac{R_1\times R_2}{R_1 + R_2}$ (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 = $R_C_D_E = \frac{2R\times 2R}{2R + 2R} =$ 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

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