Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.99 A. When the resistors are connected in parallel to the battery, the total current from the battery is 10.7 A. Determine the two resistances.
To determine the two resistances, let's start by using Ohm's law to calculate the resistance values in the two given scenarios.
1. Series Connection:
When two resistors are connected in series, the total resistance (R_total) is the sum of the individual resistors' values.
Based on Ohm's law (V = I * R), where V is the voltage, I is the current, and R is the resistance, we can derive the equation for total resistance in a series circuit as follows:
R_total = R(smaller) + R(larger)
We know that the current (I) is 1.99 A and the voltage (V) is 12.0 V.
Rearranging the equation to solve for the total resistance, we have:
R_total = V / I
R_total = 12.0 V / 1.99 A
Calculating this, the total resistance in the series circuit is approximately 6.03 Ω.
2. Parallel Connection:
When two resistors are connected in parallel, the inverse of the total resistance (1/R_total) is equal to the sum of the inverses of the individual resistances.
Using the same equation (V = I * R), we get:
R(smaller) = V / I
R(larger) = V / I
Given that the current (I) is 10.7 A and the voltage (V) is 12.0 V, we can solve for the individual resistances:
1/R_total = 1/R(smaller) + 1/R(larger)
1/R_total = 1 / (V / I) + 1 / (V / I)
1/R_total = 1 / (12.0 V / 10.7 A) + 1 / (12.0 V / 10.7 A)
After evaluating this expression, we find that the inverse of the total resistance (1/R_total) is approximately 1.68 S (Siemens).
Finally, to find the individual resistance values, we can take the reciprocals of the inverse total resistance:
R(smaller) = 1 / (1.68 S)
R(larger) = 1 / (1.68 S)
Evaluating these equations, we get the individual resistances approximately 0.595 Ω for the smaller resistor and 0.595 Ω for the larger resistor.
Therefore, the two resistance values are 0.595 Ω (smaller) and 0.595 Ω (larger) when rounded to three decimal places.
R(smaller) =r
R(larger) = R
in series:
1.99(R+r)=12 . ... ..(1)
in parallel:
1/R +1/r=1/R₀
R₀=rR/(R+r)
10.7 R₀=12
10.7•rR/(R+r)=12 …..(2)
Multiply (1) by (2)
1.99• (R+r)•10.7•rR/(R=r) =144
r=6.76/R …… (3)
From (1)
R+r=6.03 ……(4)
Substitute (30 in (4)
6.76/R + R =6.03
R²-6.03R+6.76 =0
R = 3.015±sqrt(9.09-6.76) =
=3.015±1.53
Two roots:
R₁=4.545 => r₁=6.76/R₁=1.49
R₂=2.485 => r₂=6.76/R₂=2.72
We take the first pair as R>r
Ans. : R₁=4.545 Ω, r₁=1.49 Ω