Equivalent Resistance with Error Calculation JEE Physics

 

❓ Question

Two resistors are connected in parallel:

$$R_1=(4\pm0.8)\Omega$$
$$R_2=(4\pm0.4)\Omega$$

Find the equivalent resistance with error.

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✍️ Solution

For parallel combination,

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

Using the mean values,

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

Therefore,

$$\boxed{R_{\text{eq}}=2\Omega}$$

For maximum error in parallel combination,

$$\Delta R_{\text{eq}} = R_{\text{eq}}^2 \left( \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2} \right)$$

Substituting the values,

$$\Delta R_{\text{eq}} = (2)^2 \left( \frac{0.8}{4^2} + \frac{0.4}{4^2} \right)$$
$$= 4\left(\frac{1.2}{16}\right)$$
$$\Delta R_{\text{eq}}=0.3\Omega$$

✅ Final Answer

$$\boxed{R_{\text{eq}}=(2\pm0.3)\Omega}$$