Question

Two resistors R₁ = (4 ± 0.8) Ω and R₂ = (4 ± 0.4) Ω are connected in parallel. The equivalent resistance of their parallel combination will be:

• A. (4 ± 0.4) Ω
• B. (2 ± 0.4) Ω
• C. (2 ± 0.3) Ω
• D. (4 ± 0.3) Ω

Answer 1

Answer: (C)

(2 ± 0.3) Ω

Answer 2

The equivalent resistance $R_{eq}$ of two resistors $R_1$ and $R_2$ connected in parallel is given by the formula:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

The given resistances are $R_1 = (4 \pm 0.8) \Omega$ and $R_2 = (4 \pm 0.4) \Omega$.
The nominal values are $R_{1,nominal} = 4 \Omega$ and $R_{2,nominal} = 4 \Omega$.
The uncertainties are $\Delta R_1 = 0.8 \Omega$ and $\Delta R_2 = 0.4 \Omega$.

First, calculate the nominal value of the equivalent resistance:

$\frac{1}{R_{eq,nominal}} = \frac{1}{R_{1,nominal}} + \frac{1}{R_{2,nominal}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$R_{eq,nominal} = 2 \Omega$.

Next, calculate the uncertainty in $R_{eq}$. The formula for parallel resistance involves the sum of inverse resistances. Let $X = \frac{1}{R}$. The uncertainty in $X$ is related to the uncertainty in $R$ by $\Delta X = \left|\frac{dX}{dR}\right| \Delta R$.

For $X = R^{-1}$, $\frac{dX}{dR} = -R^{-2}$. So, $\Delta X = |-R^{-2}| \Delta R = \frac{\Delta R}{R^2}$.

Let $X_{eq} = \frac{1}{R_{eq}}$, $X_1 = \frac{1}{R_1}$, and $X_2 = \frac{1}{R_2}$.
The relationship is $X_{eq} = X_1 + X_2$.
When quantities are added, their absolute uncertainties add up (for maximum possible error):
$\Delta X_{eq} = \Delta X_1 + \Delta X_2$.

Calculate $\Delta X_1$ and $\Delta X_2$ using the nominal values of $R_1$ and $R_2$:

$\Delta X_1 = \frac{\Delta R_1}{R_{1,nominal}^2} = \frac{0.8}{4^2} = \frac{0.8}{16} = 0.05 \Omega^{-1}$.
$\Delta X_2 = \frac{\Delta R_2}{R_{2,nominal}^2} = \frac{0.4}{4^2} = \frac{0.4}{16} = 0.025 \Omega^{-1}$.

Now, calculate $\Delta X_{eq}$:

$\Delta X_{eq} = 0.05 + 0.025 = 0.075 \Omega^{-1}$.

Finally, relate $\Delta X_{eq}$ back to $\Delta R_{eq}$. We have $R_{eq} = X_{eq}^{-1}$.
The uncertainty in $R_{eq}$ is $\Delta R_{eq} = \left|\frac{dR_{eq}}{dX_{eq}}\right| \Delta X_{eq}$.
For $R_{eq} = X_{eq}^{-1}$, $\frac{dR_{eq}}{dX_{eq}} = -X_{eq}^{-2}$.
So, $\Delta R_{eq} = |-X_{eq}^{-2}| \Delta X_{eq} = \frac{\Delta X_{eq}}{X_{eq}^2}$.
Using the nominal value $X_{eq,nominal} = \frac{1}{R_{eq,nominal}} = \frac{1}{2} = 0.5 \Omega^{-1}$:

$\Delta R_{eq} = \frac{0.075}{(0.5)^2} = \frac{0.075}{0.25} = \frac{0.075 \times 4}{0.25 \times 4} = \frac{0.3}{1} = 0.3 \Omega$.

The equivalent resistance is $R_{eq} = R_{eq,nominal} \pm \Delta R_{eq} = (2 \pm 0.3) \Omega$.