📂Physics

Finding the Equivalent Resistance of Parallel Circuits Easily

Buildup

Imagine calculating the equivalent resistance of the circuit shown above. Of course, if we change it into a parallel circuit like below, we can find the answer through the formula.

When there are $n$ resistors, the formula for resistance in parallel is $\displaystyle {{1} \over {R}} = {{1} \over {R_{1}}} + {{1} \over {R_{2}}} + \cdots + {{1} \over {R_{n}}}$. Substituting the resistors into the formula gives us

$$\begin{align*} {{1} \over {R}} =& {{1} \over {2}} + {{1} \over {5}} + {{1} \over {5}} \\ =& {{1} \over {2}} + {{2} \over {5}} \\ =& {{5} \over {10}} + {{4} \over {10}} \\ =& {{9} \over {10}} \end{align*}$$

Therefore, the equivalent resistance is the reciprocal of $1/R$, which is $\displaystyle R = {{10} \over {9}}$.

The problem is that this standard solution is unexpectedly difficult and involves a lot of calculations. It’s almost obvious because it’s the addition of fractions, so there are a lot of minor multiplications and the results also come out in reciprocal form during the process of finding a common denominator. If you see a mistake in a resistance problem, it’s usually because you either made an unbelievable mistake in addition or forgot to take the reciprocal at the end. As physics progresses with the grade level, it becomes a race of how quickly you can calculate within a given time, and thus parallel resistance can inevitably become a burden. It’s not that you get it wrong because you don’t know the equivalent resistance; it’s because you rush due to impatience, leading to more mistakes. Here, a method is introduced that can solve it easier and quicker, or at least with fewer mistakes. This method can be used primarily because most resistors are given as natural numbers in high school and below. It might not work everywhere, but that’s exactly why it decisively solves the problems that can be solved.

Tips

The least common multiple of $2$ and $5$ is $10$, and $2 \Omega$ can be seen as $10 \Omega$ ohm resistors connected in parallel $5$ times. Similarly, $5 \Omega$ can be seen as $10 \Omega$ ohm resistors connected in parallel $2$ times. Since both $2 \Omega$ and $5 \Omega$ are changed to $10 \Omega$, the calculation ends simply by dividing by the number of $10 \Omega$. In fact, the number of $10 \Omega$ is $9$, and the equivalent resistance calculated in this manner is $\displaystyle {{10} \over {9}}$. As the problem of calculating fractions turns into the problem of finding the least common multiple and simply counting numbers, the solution becomes faster and checking the answer is easier. Strictly speaking, this kind of calculation is already included in the parallel resistance formula, but a few minor steps are skipped to increase speed.

Examples

[1]

Calculate the parallel resistance of $2 \Omega$, $3 \Omega$, $7 \Omega$.

Solution

• The least common multiple of the resistors is $42 \Omega$.
• $2 \Omega$, $3 \Omega$, $7 \Omega$ are equivalent to having $42 \Omega$ parallel-connected $21$ times, $14$ times, $6$ times, respectively.
• Therefore, the parallel resistance is $$\displaystyle R = {{42} \over {21 +14 +6 }} = {{42} \over {41}}$$

[2]

Calculate the parallel resistance of $3 \Omega$, $6 \Omega$, $10 \Omega$.

Solution

• The least common multiple of the resistors is $30 \Omega$.
• $3 \Omega$, $6 \Omega$, $10 \Omega$ are equivalent to having $30 \Omega$ parallel-connected $10$ times, $5$ times, $3$ times, respectively.
• Therefore, the parallel resistance is $$\displaystyle R = {{30} \over {10 +5 + 3 }} = {{5} \over {3}}$$