Resources | Subject Notes | Physics
To calculate the combined resistance of two resistors connected in parallel.
In a series circuit, components are connected end-to-end, so the current is the same through each component, but the total resistance increases. In a parallel circuit, components are connected side-by-side, so the voltage across each component is the same, and the total resistance decreases.
When resistors are connected in parallel, they provide multiple paths for the current to flow. This results in a lower overall resistance compared to having the same resistors in series.
The combined resistance ($R_{eq}$) of two resistors in parallel is calculated using the following formula:
$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} $
Where:
To find the equivalent resistance ($R_{eq}$), you can follow these steps:
Consider two resistors connected in parallel with resistances $R_1 = 10 \, \Omega$ and $R_2 = 20 \, \Omega$. Calculate the combined resistance ($R_{eq}$).
Step | Calculation | Result |
---|---|---|
Calculate reciprocal of $R_1$ | $ \frac{1}{R_1} = \frac{1}{10 \, \Omega} = 0.1 \, \Omega^{-1} $ | |
Calculate reciprocal of $R_2$ | $ \frac{1}{R_2} = \frac{1}{20 \, \Omega} = 0.05 \, \Omega^{-1} $ | |
Add the reciprocals | $ \frac{1}{R_{eq}} = \frac{1}{10 \, \Omega} + \frac{1}{20 \, \Omega} = 0.1 \, \Omega^{-1} + 0.05 \, \Omega^{-1} = 0.15 \, \Omega^{-1} $ | |
Take the reciprocal of the sum | $ R_{eq} = \frac{1}{0.15 \, \Omega^{-1}} = \frac{1}{0.15} \, \Omega = \frac{100}{15} \, \Omega = \frac{20}{3} \, \Omega \approx 6.67 \, \Omega $ |
Therefore, the combined resistance of the two resistors in parallel is approximately $6.67 \, \Omega$.