Question

When two unknown resistors are connected in series with a battery, the battery delivers 230 W and carries a total current of 5.00 A. For the same total current, 55.0 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors.

Answer 1

Solution:-
Let us assume the unknown resistors be R1 andR2.When the resistors are connected in series then the equivalent resistance is
R = R1 + R2 ---------(1)
The battery delivers a power P = 230W
The battery carries a total current of I = 5.00A
We know from the relation
P = I2 * R
or R = (P/I2) = [(230)/(5.00)2] =9.2
From equation (1),we get
R1 + R2 = 9.2 ---------(2)
When the resistors are connected in parallel,then theequivalent resistance is
1/R = 1/R1 + 1/R2
or R = [(R1R2)/(R1 +R2)]
When the resistors are connected in parallel,the powerdelivered to the resistors is P = 50.0W and the current is I =5.00A,therefore,we get
P = I2 * R
or [(R1R2)/(R1 +R2)] = (P/I2) = (55.0/(5.00)2) =2.2
or R1R2 = 2 * (R1 +R2) ---------(3)
From equations (2) and (3),we get
R1R2 = 2 * 9.2 = 18.4
or R1= (18.4/R2) ----------(4)
Substituting the value of R1 in equation (2),weget
(18.4/R2) + R2 = 9.2
or R2^2 - 9.2R2 + 18.4 = 0
Solving the above quadratic equation,we get
R2 = 6.26 Ω or R2 = 2.93 Ω
When R2 = 6 Ω,then from equation (2),weget
R1 + 6.26 = 9.2
or R1 = 2.94 Ω
Therefore,the value of the two resistors are R1 =2.94Ω and R2 = 6.26 Ω.