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  • Let there be n resistors R_1 ............R_n with R_max = max (R_1......... R_n) and R_min = min {R_1 ..... R_n}.

Let there be n resistors R_1 ............R_n with R_{max} = max (R_1......... R_n) and R_{min} = min (R_1 ..... R_n). Show that when they are connected in parallel, the resultant resistance RP< Rmin and when they are connected in series, the resultant resistance R_S > R_{max}. Interpret the result physically.

Answers (1)

 

\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_{2}}+.....+\frac{1}{R_n}

\frac{R_{min}}{R_{parallel}}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}....+\frac{R_{min}}{R_n}
\frac{R_{min}}{R_{parallel}}> 1

R_{series}= R_1+R_2...+R_n

R_{series}>R_{max}

This can be represented physically as shown below:

Posted by

infoexpert24

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