Physics, asked by 248064379, 7 months ago

What is the (a) highest (b) lowest total resistance, that can be secured by combinations of four coils of resistances 4 ohm, 8 ohm, 12 ohm and 24 ohm​

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Answered by BrainlyTwinklingstar
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AnSwer :-

(a) the highest resistance is secured by combining all four coils of resistance in series

\bigstarThe combined resistance of any number of resistance connected in series is equal to the sum of the individual resistance .i.e.,

➠ R{\sf {_s}} = R₁ + R₂ + R{\sf _{3}} + R{\sf _{4}}

➠ R{\sf {_s}} = 4Ω + 8Ω + 12Ω + 24Ω

➠ R{\sf {_s}} = 28Ω

(b) the lowest is secured by combining all four coils of resistance in parallel

\bigstar the reciprocal of combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistance .i.e.,

{\leadsto { \sf  \dfrac{1}{R_{p}}  =  \dfrac{1}{R_{1}}  +  \dfrac{1}{R_{2}} +  \dfrac{1}{R_{3}}  +  \dfrac{1}{R_{4}}  }}

{\leadsto { \sf  \dfrac{1}{R_{p}}  =\dfrac{1}{4}  +  \dfrac{1}{8}  + \dfrac{1}{12}  +  \dfrac{1}{24}  }}

{\leadsto { \sf  \dfrac{1}{R_{p}}  =  \dfrac{6 + 3 + 2 + 1}{24}    }}

{\leadsto { \sf  \dfrac{1}{R_{p}}  =  \dfrac{12}{24}   =  \dfrac{1}{2}  }}

 { \leadsto { \sf R_p = 2Ω }}

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#sanvi.

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