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Equivalent resistance that can not be obtained by the combination of three resistance of four ohms each is

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1.33 Ω

6 Ω

3 Ω

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Detailed Solution

Equivalent resistance that can not be obtained by the combination of three resistance of four ohms each is

6 Ω .

For resistances in series (Say

R_{1}, R_{2} R_{3}

) the final value,

X = (R_{1} + R_{2} + R_{#})

Ohms.
For parallel connection, the formula is

\frac{1}{X} = (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{#}})

.
There can be 4 alternatives-
1. All three Resistances (R, S & T) in series, and the value would be

(R_{1} + R_{2} + R_{3}) = 12

Ohms.

2. All three in parallel and the final value would be,
Question Image

\frac{1}{X} = (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}})

\frac{1}{X} = (\frac{1}{4} + \frac{1}{4} + \frac{1}{4})

\frac{1}{X} = (\frac{3 + 3 + 3}{12})

\frac{1}{X} = \frac{9}{12} = \frac{3}{4}

X = \frac{4}{3} = 1.33

ohm
3. Two resistances

(R_{!} + R_{3})

in series and a third

(R_{2})

in parallel to them.

R_{f} = (R_{1} + R_{3})

\Rightarrow R_{f} = 4 + 4

\Rightarrow R_{f} = 8

\Rightarrow \frac{1}{R^{'}} = \frac{1}{R_{F}} + \frac{1}{R_{3}}

\Rightarrow \frac{1}{R^{'}}

\frac{1}{8} + \frac{1}{4}

\Rightarrow \frac{1}{R^{'}} = \frac{1 + 2}{8}

\Rightarrow \frac{1}{R^{'}} = \frac{8}{3}

\Rightarrow \frac{1}{R^{'}} = 2.667

The final value would be two resistors in series and one in parallel = 2.67 Ohms
4. Two Resistances(R+S)in parallel, in series with the third(T).

\frac{1}{R^{'}} = \frac{1}{R_{0}} + \frac{1}{R_{1}}

\Rightarrow \frac{1}{R^{'}} = \frac{1}{4} + \frac{1}{4}

\Rightarrow \frac{1}{R^{'}} = \frac{1 + 1}{4}

\Rightarrow \frac{1}{R^{'}} = \frac{2}{4}

\Rightarrow R^{'} = 2

\Rightarrow 2 + 4 = 6

The final value would be two resistors in series and one in parallel = 6 Ohms

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