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How can three resistors of resistances $of 3 ohm, 3 ohm, 3 ohm$ be connected to give a total resistance of $a) 9 ohm b) 1 ohm$ .

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9 Ω : 1 Ω

4.5 Ω; 1 Ω

7.5 Ω; 1 Ω

None of the above

answer is B.

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Detailed Solution

The total resistance of the circuit is

4.5 Ω .

The net equivalent resistance of the circuit is

1 Ω .

Given:-

R₁ = 3 Ω

R₂ = 3 Ω

R₃ = 3 Ω

(a) When

R_{2}

and

R_{3}

are connected in parallel with

R_{1}

in series we get ⇒

\frac{1}{R} = \frac{1}{R_{2}} + \frac{1}{R_{3}}

⇒

\frac{1}{R} = \frac{1}{3} + \frac{1}{3}

⇒

\frac{1}{R} = \frac{1 + 1}{3}

\frac{1}{R} = \frac{2}{3}

⇒

2 R = 3 Ω

⇒ R =

\frac{3}{2}

Ω Resistance in series

= R + R₁

Resistance in series =

\frac{3}{2}

+ 3

= \frac{3 + 6}{2} = \frac{9}{2}

= 4.5 Ω

Hence, the total resistance of the circuit is 4.5 Ω. (b) When

R_{1}

R_{2}

, and

R_{3}

are connected in parallel we get,

\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

⇒

\frac{1}{R} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}

⇒

\frac{1}{R} = \frac{1 + 1 + 1}{3}

\frac{1}{R} = \frac{3}{3}

\frac{1}{R} = 1

Hence, the net equivalent resistance of the circuit is

1 Ω .

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