When two resistors RA and RB are combined in parallel and connected to a battery it is observed that the ratio of the power dissipated in RA to that dissipated in RB is equal to 4.0. If this arrangement is dismantled and the same two resistors connected in series to the same battery, what is the ratio of the power dissipated in RA to that dissipated in RB in this new circuit?

When resistors are connected in parallel, the power dissipated in each resistor is given by: r2 R - P = where V' is the voltage across the resistor and R is the resistance of the resistor. Given that the ratio of the power dissipated in I 4 to that dissipated in 3 when they are connected in parallel is 4, we can denote the power dissipated in Ry as Py andin R as Pp: 4 e Now, let's analyze the situation when they are connected in series. When resistors are connected in series, the total resistance is the sum of the individual resistances. Thus, for resistors K 4 and R connected in series, the total resistance is Recpjes = Ry + Rpg. The total power dissipated in the series circuit is given by: & Pseries & SOCICs Given that the same battery is used in both cases, the voltage V' remains the same. Now, let's find the ratio of the power dissipated in I 4 to that dissipated in {3 when they are connected in series: T I 54515 I B =)= o \\ From the given information, we know that % = 4 so: = 4 o A Now, let's rearrange this equation to find the ratio of the power dissipated in /7 4 to that dissipated in B when they are connected in series: o 1 B 1 1 b So,when I? 4 and R are connected in series, the ratio of the power dissipated in I 4 to that dissipatedin Rp is