The equivalent resistance of four resistors joined in parallel is 20 ohms. The current flowing through them are 0.6, 0.3, 0.2 and 0.1 ampere. The value of each resistor is:
When multiple resistors are connected in a parallel circuit, two fundamental principles apply:
We are given the equivalent resistance (\(R_{eq}\)) of four resistors in parallel as \(20 \text{ ohms}\) and the current flowing through each resistor. Our goal is to find the value of each individual resistor.
The first step is to determine the total current (\(I_{total}\)) flowing through the parallel circuit. We can find this by summing the individual currents given:
Using the principle of current division in parallel circuits:
\[ I_{total} = I_1 + I_2 + I_3 + I_4 \] \[ I_{total} = 0.6 \text{ A} + 0.3 \text{ A} + 0.2 \text{ A} + 0.1 \text{ A} \] \[ I_{total} = 1.2 \text{ A} \]
Next, we can determine the voltage (\(V\)) across the parallel combination. Since the voltage is the same across all resistors in parallel, finding this voltage will allow us to calculate the individual resistances. We use Ohm's Law, which states \(V = I \times R\). Here, we'll use the total current and the equivalent resistance:
\[ V = I_{total} \times R_{eq} \] \[ V = 1.2 \text{ A} \times 20 \text{ ohms} \] \[ V = 24 \text{ V} \]
So, the voltage across each of the four resistors is \(24 \text{ V}\).
With the voltage across each resistor known, and the current through each resistor given, we can now apply Ohm's Law (\(R = V/I\)) to calculate the resistance of each individual resistor:
| Resistor | Current (\(I\)) | Voltage (\(V\)) | Resistance (\(R = V/I\)) |
|---|---|---|---|
| First Resistor (\(R_1\)) | \(0.6 \text{ A}\) | \(24 \text{ V}\) | \(R_1 = \frac{24 \text{ V}}{0.6 \text{ A}} = 40 \text{ ohms}\) |
| Second Resistor (\(R_2\)) | \(0.3 \text{ A}\) | \(24 \text{ V}\) | \(R_2 = \frac{24 \text{ V}}{0.3 \text{ A}} = 80 \text{ ohms}\) |
| Third Resistor (\(R_3\)) | \(0.2 \text{ A}\) | \(24 \text{ V}\) | \(R_3 = \frac{24 \text{ V}}{0.2 \text{ A}} = 120 \text{ ohms}\) |
| Fourth Resistor (\(R_4\)) | \(0.1 \text{ A}\) | \(24 \text{ V}\) | \(R_4 = \frac{24 \text{ V}}{0.1 \text{ A}} = 240 \text{ ohms}\) |
Therefore, the values of the four resistors are \(40 \text{ ohms}\), \(80 \text{ ohms}\), \(120 \text{ ohms}\), and \(240 \text{ ohms}\).
As a final check, we can verify if these calculated individual resistances yield the given equivalent resistance of \(20 \text{ ohms}\) using the formula for equivalent resistance in parallel:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \] \[ \frac{1}{R_{eq}} = \frac{1}{40} + \frac{1}{80} + \frac{1}{120} + \frac{1}{240} \]
To sum these fractions, we find a common denominator, which is \(240\):
\[ \frac{1}{R_{eq}} = \frac{6}{240} + \frac{3}{240} + \frac{2}{240} + \frac{1}{240} \] \[ \frac{1}{R_{eq}} = \frac{6 + 3 + 2 + 1}{240} \] \[ \frac{1}{R_{eq}} = \frac{12}{240} \] \[ \frac{1}{R_{eq}} = \frac{1}{20} \] \[ R_{eq} = 20 \text{ ohms} \]
This matches the equivalent resistance given in the question, confirming our calculations for the individual resistor values are correct.
The values of the resistors are \(40 \text{ ohms}\), \(80 \text{ ohms}\), \(120 \text{ ohms}\), and \(240 \text{ ohms}\).
Three resisters of 3 ohm, 10 ohm and 15 ohm are connected in parallel in a 30 V circuit. The current will that flow through the 3-ohm resistor is:
In series–parallel combination of resistance, the minimum number of resistance required is _____.
If n identical resistance, each of resistance R, are connected in parallel, the equivalent resistance is:
Four 100 Ω resistors are connected in parallel. The equivalent resistance of the parallel connection is: