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Explain series connection of resistors. Derive equation of equivalent resistance (R_(S)). |
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Answer» Solution :When more than one resistors are CONNECTED between two POINT in such a way that current FLOWING through each resistor is equal (same) then such connection of resistors is called series connection. In series connection sum of p.d. (voltage) across each resistor is equal to total voltage supplied (emf of cell). `RARR` As shown in figure two resistor `R_(1) and R_(2)` are connected between point A and B ,  ` rArr` Current flowing through both resistors is same (I) . `therefore `p.d. across resistor `R_(1)`. p.d. across `R_(2)`, `V_(2) = IR_(2) ""` ....(2) V = terminal voltage of battery ` V = V_(1) + V_(2)` = `IR_(1) + IR_(2)` `V = I (R_(1) + R_(2))` `I = I_(1) + I_(2)` ` = (V)/(R_(1)) + (V)/(R_(2))` = I = V`((1)/(R_(1)) + (1)/(R_(2) ) )` `(I)/(V) = (1)/(R_(1)) + (1)/(R_(2))` `(V)/(I) = R_(eq)` `(I)/(V) = (1)/(R_(eq))` `(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))` `rArr` In parallel connection RECIPROCAL equivalent resistance is equal to sum of reciprocal of individual resistors. `rArr` In parallel connection of resistors equivalent resistance is smaller than smallest value of resistance. `rArr R_(1), R_(2), .... , R_(n)` resistors are connected inparallel then `R_(eq) = R_(p)`, `(1)/(R_(eq))= (1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + .... + (1)/(R_(n))` `rArr " If " R_(1) = R_(2) ...., R_(n) = R,` then equivalent resistance, `(1)/(R_(eq)) = (1)/(R_(p)) = (1)/(R) + (1)/(R) + .... + ` n times. `rArr` In such connection sum of current flowing through each resistor is equal to total current in the circuit.  `rArr " Let " R_(1) and R_(2)` resistance are connected between a and b and battery of terminal voltage V is connected. `rArr` Let current in the circuit is I. It is divided into two branch `I_(1) and I_(2)` at point a. `rArr` Let current in `R_(1) " be" I_(1)` and current in `R_(2) " be " I_(2)` at point a, `I = I_(1) + I_(2) ` .... (1) `rArr` By Ohm.s law , p.d . across `R_(1)`, V = `I_(1) R_(1)` `I_(1) = (V)/(R_(1)) "" `... (2) `rArr`p.d. across `R_(2)` , `V = I_(2) R_(2)` `I_(2) = (V)/(R_(2)) `...(3) `(1)/(R_(eq)) = (n)/(R) ` `R_(eq) = (R)/(n)` |
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