Saved Bookmarks
| 1. |
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 R_(p) = R_("min")and when they are connected in series, the resultant resistance R_(s) gt R_("max") . Interpret the result physically. |
Answer» Solution :Let some minimum value of resistance be `R_(" min")`  Let `R_("max") and R_("min")` be maximum and minimum resistance. For PARALLEL connection, `(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + .... + (1)/(R_("min")) + ... + (1)/(R_(n))` By multiplying by `R_("min") ` on both side, `(R_("min"))/(R_(P)) = (R_("min"))/(R_(1)) + (R_("min"))/(R_(2))+ ..... + (R_("min"))/(R_(n))` `(R_("min"))/(R_(P)) gt ` 1 (Right side value is more than 1) `therefore(R_(p))/(R_("min")) lt 1 ` `therefore R_(p) lt R_("min")`  Series connection : From `R_(1), R_(2), ...., R_(n)R_("max") ` be maximum value, For series connection, `R_(S) = R_(1) + R_(2) + .... + R_("max") +.... + R_(n)` `R_(s) = R_("max") + (R_(1) +.... + R_(n))` `R_(s) gt R_("max")` ( Value of summation of resistance other than`R_("max") ` is POSITIVE.  For parallel connection : From FIGURE (a) `R_("min")` resistance is minimum resistance for loop given in figure (b). But in figure (b) there are (n- 1) loop so additional resistor (n - 1) . HENCE. current obtained in figure (b) is larger that current obtained in figure (a). `therefore ` Equivalent resistance of figure (b) will be LESS than `R_("min")` `therefore R_(p) lt R_("min")` For series connection : In figure (c) provide maximum resistance `R_("max")` Thus, current in figure (d) will be less than current in figure (c). thus, equivalent resistance of figure (d) `ltR_("max") therefore R_(S) gt R_("max")` 
|
|