three resistors with values R1,R2 and R3, respectively are connected in series to a 6-v battery.the total current flowing through the circuit is I=0.6mA.when these three resistors are connected in parallel to a 6v battery,the total current is I=6.2mA. also the current flowing through R2 is I=2mA. determing the individual resistance of the three resistors.note R1<R2<R3
we know that
6/R2 = .002, so R2 = 3KΩ
R1+R2+R3 = 6/.0006 = 10KΩ
So, now we have
R1+R3 = 10K-3K = 7KΩ
6/R1 + 6/R3 = .0062-.002 = .0042
Now just solve for R1 and R3, remembering that R1 < R3
To find the individual resistance of the three resistors, we can use the formulas for resistors in series and parallel circuits.
When the resistors are connected in series:
In a series circuit, the total resistance (R_total) is the sum of the individual resistances (R1 + R2 + R3).
Given that the total current (I) is 0.6mA, we can use Ohm's Law (I = V/R) to find the total resistance.
R_total = V/I = 6V / 0.6mA (converting mA to A) = 10kΩ
When the resistors are connected in parallel:
In a parallel circuit, the reciprocal of the total resistance (1/R_total) is equal to the sum of the reciprocals of the individual resistances (1/R1 + 1/R2 + 1/R3).
Given that the total current (I) is 6.2mA, we can use Ohm's Law to find the total resistance.
R_total = V/I = 6V / 6.2mA (converting mA to A) = 967.7Ω
Now, we need to determine the individual resistances:
Since R_total for the series circuit is 10kΩ and for the parallel circuit is 967.7Ω, we can set up the following equations:
R1 + R2 + R3 = 10kΩ (Equation 1)
1/R1 + 1/R2 + 1/R3 = 1/R_total (Equation 2)
From Equation 2, we know that 1/R_total = 1/967.7Ω.
Solve Equation 1 for R3:
R3 = 10kΩ - R1 - R2
Substitute this expression for R3 in Equation 2:
1/R1 + 1/R2 + 1/(10kΩ - R1 - R2) = 1/967.7Ω
Using the given information that the current through R2 is 2mA, we can write:
I = V/R2
2mA = 6V / R2
R2 = 6V / 2mA = 3kΩ
Substitute this value for R2 in the equation:
1/R1 + 1/3kΩ + 1/(10kΩ - R1 - 3kΩ) = 1/967.7Ω
Now, you can solve this equation for R1 and R3.
To determine the individual resistances of the three resistors, we can use the given information about the total current and the current flowing through R2.
In a series circuit, the total resistance is equal to the sum of the individual resistances.
For the series circuit with a total current of I = 0.6mA, we can use Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance.
So, in the series circuit, we have:
V = I * R_total
6V = 0.6mA * R_total
R_total = 6V / 0.6mA
R_total = 10kΩ (kiloohms)
Since the resistors are connected in series, the total resistance is equal to the sum of the individual resistances:
R_total = R1 + R2 + R3
We are also given the current flowing through R2 as I = 2mA. Since the current is the same throughout a series circuit, we can say that the current flowing through the entire series circuit is equal to the current flowing through R2:
I = I_R2
0.6mA = 2mA
Now, let's substitute the values into the equation for the total resistance:
10kΩ = R1 + R2 + R3
To solve for the individual resistances, we need another equation. We can use the fact that when the resistors are connected in parallel, the total current is given as I = 6.2mA.
In a parallel circuit, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances:
1/R_total = 1/R1 + 1/R2 + 1/R3
Plugging in the values:
1/10kΩ = 1/R1 + 1/R2 + 1/R3
(1/10kΩ) - (1/R2) = 1/R1 + 1/R3
Now, we can substitute the known value for R2 (which is R2 = 2mA / 0.6mA = 3.33kΩ) into the equation:
(1/10kΩ) - (1/3.33kΩ) = 1/R1 + 1/R3
To calculate R1 and R3, let's solve for the common denominator of the fractions:
(3.33kΩ - 10kΩ) / (3.33kΩ * 10kΩ) = 1/R1 + 1/R3
Simplifying the equation:
(-6.67kΩ) / (3.33kΩ * 10kΩ) = 1/R1 + 1/R3
-0.2 = 1/R1 + 1/R3
Now, we have two equations:
10kΩ = R1 + 3.33kΩ + R3
-0.2 = 1/R1 + 1/R3
We can solve this system of equations to find the values of R1 and R3.