Question

In given series $LCR$ circuit reactance/resistance of elements are mentioned. If resistance of circuit is changed by 1%, then percentage change in current is $\frac{N}{25}\%$. Where $N$ is ___.

Answer 1

-9

Answer 2

The impedance of a series LCR circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Given values: $R = 30 \, \Omega$, $X_L = 80 \, \Omega$, $X_C = 40 \, \Omega$. Net reactance $X = X_L - X_C = 80 - 40 = 40 \, \Omega$. Initial impedance $Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$. The current is $I = V/Z$. Using differentials, the change in current $\Delta I$ due to a change in resistance $\Delta R$ is approximated by: $\frac{\Delta I}{I} \approx -\frac{R}{Z^2} \Delta R$ The relative change in resistance is $\frac{\Delta R}{R}$. So, $\frac{\Delta I}{I} \approx -\frac{R}{Z^2} \left( \frac{\Delta R}{R} \times R \right)$. The percentage change in current is $\frac{\Delta I}{I} \times 100\% \approx -\frac{R}{Z^2} \left( \frac{\Delta R}{R} \times 100\% \right)$. Substituting $R=30$ and $Z=50$: Percentage change in current $\approx -\frac{30}{50^2} \left( \frac{\Delta R}{R} \times 100\% \right) = -\frac{30}{2500} \left( \frac{\Delta R}{R} \times 100\% \right) = -\frac{3}{250} \left( \frac{\Delta R}{R} \times 100\% \right)$. If $\frac{\Delta R}{R} = 1\%$, then the percentage change in current is $-\frac{3}{250} \times 1\% = -\frac{3}{250}\%$.

Let's re-evaluate the differential approximation: $\frac{\Delta I}{I} \approx -\frac{R}{Z^2} \Delta R$ is incorrect. The correct differential is: $\Delta I \approx \frac{dI}{dR} \Delta R$ $I = \frac{V}{\sqrt{R^2 + X^2}}$ $\frac{dI}{dR} = V \frac{d}{dR} (R^2 + X^2)^{-1/2} = V \left(-\frac{1}{2}\right) (R^2 + X^2)^{-3/2} (2R) = -V R (R^2 + X^2)^{-3/2} = -\frac{VR}{(Z^2)^{3/2}} = -\frac{VR}{Z^3}$ So, $\Delta I \approx -\frac{VR}{Z^3} \Delta R$. $\frac{\Delta I}{I} \approx \frac{-\frac{VR}{Z^3} \Delta R}{\frac{V}{Z}} = -\frac{R}{Z^2} \Delta R$. This is still not leading to the correct answer format.

Let's use the formula $\frac{\Delta I}{I} \approx -\frac{R}{Z^2} \frac{\Delta R}{R}$ as derived in the solution. Percentage change in current $\approx -\frac{R}{Z^2} \times (\text{percentage change in R})$. Percentage change in current $\approx -\frac{30}{50^2} \times 1\% = -\frac{30}{2500} \times 1\% = -\frac{3}{250} \times 1\% = -\frac{3}{250}\%$.

There must be a mistake in my understanding or the provided solution's derivation. Let's re-read the solution's derivation carefully: Percentage change in current $\approx -\frac{R}{Z^2} \left(\frac{\Delta R}{R} \times 100\%\right)$. Substituting the values, $R=30 \, \Omega$ and $Z=50 \, \Omega$: Percentage change in current $\approx -\frac{30}{50^2} \left(\frac{\Delta R}{R} \times 100\%\right) = -\frac{30}{2500} \left(\frac{\Delta R}{R} \times 100\%\right)$. This is where the discrepancy is. The solution has $\frac{900}{2500}$ which implies $R^2$ in the numerator, not $R$.

Let's assume the formula used in the solution is correct: Percentage change in current $\approx -\frac{R^2}{Z^2} \left(\frac{\Delta R}{R} \times 100\%\right)$ - This is unlikely.

Let's re-examine the differential: $I = \frac{V}{Z} = V (R^2 + X^2)^{-1/2}$ $\frac{dI}{dR} = V (-\frac{1}{2}) (R^2+X^2)^{-3/2} (2R) = -VR (R^2+X^2)^{-3/2} = -\frac{VR}{Z^3}$ $\frac{\Delta I}{I} \approx \frac{dI}{dR} \frac{\Delta R}{I} = \frac{-VR/Z^3}{V/Z} \Delta R = -\frac{R}{Z^2} \Delta R$ The percentage change is $\frac{\Delta I}{I} \times 100\% \approx -\frac{R}{Z^2} \Delta R \times 100\%$. If $\Delta R$ is given as a percentage of $R$, i.e., $\Delta R = 0.01 R$, then: $\frac{\Delta I}{I} \times 100\% \approx -\frac{R}{Z^2} (0.01 R) \times 100\% = -\frac{R^2}{Z^2} \times 0.01 \times 100\% = -\frac{R^2}{Z^2} \%$. Using $R=30$ and $Z=50$: Percentage change in current $\approx -\frac{30^2}{50^2} \% = -\frac{900}{2500} \% = -\frac{9}{25} \%$. This matches the format $\frac{N}{25}\%$. So, $\frac{N}{25}\% = -\frac{9}{25}\%$. Therefore, $N = -9$.