Q.Find the capacitance of a capacitor which, when put in series with a 10 $\Omega$ resistor, makes the power factor equal to 0.5. Assume an 80 V-100 Hz AC supply.

[!TLDR] $\cos\phi=0.5\Rightarrow\phi=60^\circ\Rightarrow\tan60^\circ=\sqrt3=X_C/R\Rightarrow X_C=10\sqrt3\approx17.32\,\Omega$; $C=1/(2\pi f X_C)=1/(2\pi(100)(17.32 …

Given R = 10 $\Omega$, desired power factor $\cos\phi=0.5$, supply 80 V-100 Hz (only R and C in series, current leading). From $\cos\phi=0.5$, $\phi=\cos^{-1}(0.5)=60^\circ$, so $\tan\phi=\tan60^\circ=\sqrt3\approx1.732$. For a series R-C circuit, $\tan\phi=\dfrac{X_C}{R}$, so $X_C=R\tan\phi=10\times1.732=17.32\,\Omega$.\n\nSince $X_C=\dfrac{1}{2\pi fC}$, solving for C: $C=\dfrac{1}{2\pi f\,X_C}=\dfrac{1}{2\times3.1416\times100\times17.32}=\dfrac{1}{10882.6}\approx9.189\times10^{-5}$ F, matching the book's $9.187\times10^{-5}$ F within rou …

Convert the given power factor to a phase angle via $\phi=\cos^{-1}(\cos\phi)$, use $\tan\phi=X_C/R$ for a series R-C circu …

Trying to use the given 80 V supply voltage in the calculation, when in fact the power factor (and hence this problem) is independent of the voltage magnit …