Question

Pls solve this problem ​

Answer 1

Explanation:

ANSWER:

• In triangle ACD

First of all the resistor of 3 ohms are in series, so adding those we get 6ohm.

Now, the resistor on line AD is parallel to the equivalent resistance calculated Earlier.

And to find out the parallel combination equivalentcy we will:

1/R=1/6 +1/6. (where R is the total Resistance of the Triangle)

Calculating those, we get R=3 ohms.

So, Line AD has resistance of 3 ohms.

• Similarly, in Triangle ADE

Resistors of Line AD & DE are in series

Equivalent Resistance = 3+3 = 6 ohms.

Again, This equivalent resistance will be parallel to the resistance on line AE.

So, here again We will take R as the equivalent Resistance.

1/R=1/6 + 1/6

R= 3 ohms.

• Again in Triangle AEF

AE & EF resitors are in parallel.

.Equivalent Resistance = 3+3 = 6 ohms.

Now in order to find Equivalent Resistance,

Resistor of AE will be parallel to the earlier one.

So, here again We will take R as the equivalent Resistance.

1/R=1/6 + 1/6

R= 3 ohms.

• In Triangle ABF

AF & FB are in series.

So, we have equivalent resistance as 6 ohms.

Now, the resistor on line AF is parallel to the equivalent resistance calculated Earlier.

And to find out the parallel combination equivalentcy we will:

1/R=1/6 +1/6. (where R is the total Resistance of the Triangle)

Calculating those, we get

R=3 ohms.

So, Total Resistance will be 3 ohms.