A 30 ohm resistor is connected in parallel with a variable resistance R and the parallel combination is then connected in series with a 5 ohm resistor, and connected across 120 volt source. Find the minimum value of R if the power taken by R is equal to the power taken by the 5 ohm resistor.
current thru the 5 ohm resistor call I.
Then the current in the varistat must be divided such that current divides:
The total current is I
the resistance across the Parallel branch is R*30/(R+30)
The voltage across the parallel branch is I*R*30/(R+30)
power in the varReistor=Vparall^2/R
so
solve:
Vparallel^2/R=I^2*5
(I*R*30)^2/R(R+30)^2=I^2*5
R*900=5*(R+30)^2
solve for R with the quadratic equation.
To determine the minimum value of R, we need to set up an equation that relates the powers taken by R and the 5 ohm resistor.
Let's first calculate the total resistance of the circuit:
1. The parallel combination of the 30 ohm resistor and the variable resistance R can be found using the formula:
(1/total resistance) = (1/30 ohm) + (1/R)
Rearranging the equation, we get:
(1/total resistance) = (R + 30) / (30R)
Simplifying further, we have:
total resistance = (30R) / (R + 30)
2. The total resistance of the circuit is the sum of the parallel combination resistance and the 5 ohm resistor:
total resistance = (30R) / (R + 30) + 5 ohm
To find the minimum value of R, we need to set up an equation equating the power taken by R to the power taken by the 5 ohm resistor.
The power dissipated by a resistor can be calculated using the formula:
power = (current)^2 * resistance
The current passing through the circuit can be calculated using Ohm's Law:
current = voltage / total resistance
Let's substitute these values and set the two powers equal to each other:
((voltage / total resistance) ^ 2) * R = ((voltage / total resistance) ^ 2) * 5
We can cancel out the voltage term from both sides, as it is the same for both resistors:
(R / total resistance) ^ 2 = (5 / total resistance) ^ 2
Simplifying further, we have:
R^2 = 5^2
R^2 = 25
Taking the positive square root of both sides, we get:
R = 5
Therefore, the minimum value of R that satisfies the condition is 5 ohms.
To find the minimum value of resistance, we need to equate the power taken by the variable resistance R with the power taken by the 5 ohm resistor.
Let's begin by analyzing the circuit. We have a 30 ohm resistor (let's call it R1) connected in parallel with the variable resistance R (let's call it R2). This combination is then connected in series with a 5 ohm resistor (let's call it R3). The entire combination is connected across a 120 volt source.
First, we need to find the equivalent resistance of the parallel combination (R1 and R2). The formula to calculate the equivalent resistance of two resistors in parallel is given by:
1/Req = 1/R1 + 1/R2
Substituting the values, we get:
1/Req = 1/30 + 1/R
To simplify this equation, we multiply through by 30R:
30R/R + 30 = 30/R + 30
30 + 30R = 30 + 30R
This equation simplifies to:
30 + 30R = 30 + 30R
As we can see, the value of Req does not depend on the value of R.
Next, we have the series combination of Req and R3, with a total voltage of 120 volts across it. In a series circuit, the total resistance (Rtotal) is simply the sum of the individual resistances.
Rtotal = Req + R3
Rtotal = Req + 5
Now, we can calculate the power (P) taken by each resistor.
The power formula is given by:
P = (V^2) / R
For the 5 ohm resistor:
P1 = (120^2) / 5
For the total resistance (Rtotal):
P2 = (120^2) / Rtotal
Since the power taken by R is equal to the power taken by R3, we can equate the two expressions:
(120^2) / 5 = (120^2) / Rtotal
Simplifying this equation, we get:
Rtotal = 5
Substituting Rtotal = Req + 5, we have:
Req + 5 = 5
Req = 0
Therefore, the minimum value of R is 0 ohms.