Figure (A) and (B) show two resistors with resistances `R_(1)` and `R_(2)` connected in parallel and in series. The battery has a terminal voltage of e.

Suppose `R_(1)` and `R_(2)` are connected in parallel .
(a) Find the power delivered to each resistor.
(b) Show that the power used by each resistor is equal to the power supplied by the battery .
Suppose `R_(1)` and `R_(2)` are now connected in series
(c ) Find teh power delivered to each resistor.
(d) Show that the sum of the power by each resistor is equal to the power supplied by the battery.
(e ) Which configuration , parallelor series, uses more power ?

Explanation

(a) When two resistors are connected in paralle, the current through each resistor is `l_(1) = (epsilon)/(R_(1)),l_(2)= (epsilon)/(R_(2))`
and the power delivered to each resistor is given by
`P_(1) = l_(1)^(2) R_(1) = (epsilon^(2))/( R_(1)), P_(2) = l_(2)^(2) R_(2) = ( epsilon^(2))/(R_(2))`
(b) The total power delivered to the two resistors is
`P_(R ) = P_(1)+ P_(2) + (epsilon^(2))/( R_(1))+ (epsilon^(2))/(R_(2)) implies (epsilon^(2))/(R_(eq))`
where,
`(1)/(R_(eq))= (1)/( R_(1))+ (1)/(R_(2))implies R_(eq)= (R_(1)R_(2))/(R_(1)+R_(2))`
is the equivalent resistance of the circuit .
On the other hand , the total power supplied by the battery is `P = i epsilon`, where `l= l_(1) - l_(2)` , as seen from the figure. Thus , `P = l_(1) epsilon + l_(2) epsilon = ((epsilon)/(R_(1)))epsilon+ ((epsilon)/(R_(2)))epsilon = (epsilon^(2))/(R_(1))+(epsilon^(2))/(R_(2))= (epsilon^(2))/(R_(eq))= P_(R ) ` as required by energy conservation.
(c ) When the two resistors are connected in series, the equivalent resistance becomes `R_(eq)^(')= R_(1) + R_(2)`
and the current through the resistors are
`l_(1) = l_(2) = l = ( epsilon)/( R_(1)+R_(2))`
Therefore, the power delivered to each resistor is
`P_(1) = l_(1)^(2) R_(1) = ((epsilon)/(R_(1)+R_(2)))^(2)R_(1),P_(2) = l_(2)^(2) R_(2) = ((epsilon)/(R_(1)+R_(2)))^(2)R_(2)`
Contrary to waht we have seen in the parallel case, when connected in series, the greater the resistance, the greater the fraction of the power delivered. Once again, if the loads are ligth bulbs, the one with greater resitance will be brighter.
(d) The total power delivered to the resistor is
`P_(R )^(')= P_(1)+P_(2) = ((epsilon)/(R_(1)+R_(2)))^(2) R_(1)+ ((epsilon)/(R_(1)+R_(2)))^(2) R_(2)= ( epsilon^(2))/(R_(1)+R_(2)) = (epsilon^(2))/(R_(eq)^('))`
On the other hand, the power supplied by the battery is
`P'= l epsilon = ((epsilon)/(R_(1)+R_(2)))epsilon= (epsilon^(2))/(R_(1)+R_(2))= (epsilon^(2))/(R_(eq)^('))`
Again, we see that `P' = P_(R ) ^(')` as required by energy conservation.
(e ) Comparing the results obtained in (b) and (d), we see that
`P _(epsilon) = (epsilon^(2))/(R_(1))+(epsilon^(2))/(R_(2)) gt (epsilon^(2))/(R_(1)+R_(2))=P_(epsilon)^(')`
which means that the parallel connection uses more power. This is true since the equivalent resistance of two resistors connected in parallel is always than that connected in series.