Equivalent Resistance Triangle Of Resistors

Iii resistor, each equal to 3 Ω are continued then every bit to form a triangle. The equivalent resistance between any two vertices of the triangle is:

one Ω
2 Ω
3 Ω
9 Ω

Reply (Detailed Solution Beneath)

Option two : two Ω

Costless

Electric charges and coulomb'due south law (Basic)

10 Questions 10 Marks 10 Mins

Concept :

Resistance:

The measurement of the opposition of the flow ofelectric current through aconductor is calledresistance of that conductor. It is denoted pastR.

There are mainlytwo ways of the combination of resistances:

i. Resistances in serial:

F1 J.K 25.3.20 pallavi D3

Whentwo or more resistances are connected i after another such that thesame electric current flows through them are chosenequally resistances in series.
Thenet resistance/equivalent resistance (R) of resistances in series is given by:
Equivalent resistance,R = R1 + R2

2. Resistances in parallel:

F1 J.K 25.3.20 pallavi D4

When theterminals of ii or more resistances are connected at the same two points andthe potential divergence across them is equal is calledresistances in parallel.
Thenet resistance/equivalent resistance(R) of resistances in parallel is given by:

\(\frac{1}{{{R_{eff}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

Calculation:

Three equal resistance of 3Ω are connected on three sides of the triangle

Let's take one side AB and find the equivalent resistance across it.

⇒ 3Ω + 3Ω = 6Ω Resistance is in parallel with 3Ω resistance

Therefore, their equivalent resistance will be -

\(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_{two}}}}\)

\(\frac{1}{{{R_{eq}}}} = \frac{one}{{6}} + \frac{one}{{iii\;}} \)

Req = 2 Ω

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