Explanations & Calculations
Consider the sketch attached
R L = 15 Ω X L = 2 π f L = 30.16 Ω cos θ = 0.866 → θ = 3 0 0 ( f o r p a r t a o n l y ) \qquad\qquad
\begin{aligned}
\small R_L&= \small 15\Omega\qquad\qquad\qquad X_L=2\pi fL=30.16\Omega\\
\small \cos\theta&=\small 0.866\to\theta =30^0 (for\,part\,a\,only)
\end{aligned} R L cos θ = 15Ω X L = 2 π f L = 30.16Ω = 0.866 → θ = 3 0 0 ( f or p a r t a o n l y )
All the resistance, inductance & capacitance are connected in series hence the same current flows through each.
a)
From the phasor diagram for part (a), V R = 117 cos θ = 117 × 0.866 = 101.32 V \qquad\qquad
\begin{aligned}
\small V_R&= \small 117\cos\theta =117\times 0.866=101.32V
\end{aligned} V R = 117 cos θ = 117 × 0.866 = 101.32 V
Then by V = i R \small V=iR V = i R , i = 101.32 V 15 Ω = 6.75 A \qquad\qquad
\begin{aligned}
\small i&= \small \frac{101.32V}{15\Omega}=\bold{6.75A}
\end{aligned} i = 15Ω 101.32 V = 6.75A
Then the voltage drop across the inductive part is, V L = i X L = 203.58 V \qquad\qquad
\begin{aligned}
\small V_L&= \small iX_L=203.58\,V
\end{aligned} V L = i X L = 203.58 V
117 sin θ = V c − V L V c = 117 sin 30 + 203.58 = 262.08 V C = 6.75 A 2 π × 60 H z × 262.08 V = 68.32 μ F \qquad\qquad
\begin{aligned}
\small 117\sin\theta&= \small V_c-V_L\\
\small V_c&= \small 117\sin30+203.58=262.08V\\
\small C&= \small \frac{6.75A}{2\pi\times 60Hz\times 262.08V}\\
&= \small \bold{68.32\,\mu F}
\end{aligned} 117 sin θ V c C = V c − V L = 117 sin 30 + 203.58 = 262.08 V = 2 π × 60 Hz × 262.08 V 6.75 A = 68.32 μ F
b)
By refering to the phasor diagram for part (b), V c o i l = V c i Z c o i l = i X c R 2 + X L 2 = X c = 1 2 π f C C = 1 2 π × 60 × 1 5 2 + 30.1 6 2 = 78.75 μ F \qquad\qquad
\begin{aligned}
\small V_{coil}&= \small V_c\\
\small iZ_{coil}&= \small iX_c\\
\small \sqrt{R^2+X_L^2}&=\small X_c=\frac{1}{2\pi f C}\\
\small C&=\small \frac{1}{2\pi \times 60\times \sqrt{15^2+30.16^2}}\\
&= \small \bold{78.75\,\mu F}
\end{aligned} V co i l i Z co i l R 2 + X L 2 C = V c = i X c = X c = 2 π f C 1 = 2 π × 60 × 1 5 2 + 30.1 6 2 1 = 78.75 μ F
Then capacitive reactance is X c = 1 2 π f C = 1 2 π × 60 × 78.75 × 1 0 − 6 = 33.68 Ω \qquad\qquad
\begin{aligned}
\small X_c&= \small\frac{1}{2\pi f C}\\
&= \small \frac{1}{2\pi\times 60\times 78.75\times10^{-6} }\\
&= \small 33.68 \,\Omega
\end{aligned} X c = 2 π f C 1 = 2 π × 60 × 78.75 × 1 0 − 6 1 = 33.68 Ω
11 7 2 = V R 2 + ( V c − V L ) 2 = i 2 [ R 2 + ( X c − X L ) 2 ] i = 117 1 5 2 + ( 33.68 − 30.16 ) 2 = 7.59 A \qquad\qquad
\begin{aligned}
\small 117^2 &= \small V_R^2+(V_c-V_L)^2\\
&=\small i^2\big[R^2+(X_c-X_L)^2\big]\\
\small i&= \small \frac{117}{\sqrt{15^2+(33.68-30.16)^2}}\\
&= \small \bold{7.59\,A}
\end{aligned} 11 7 2 i = V R 2 + ( V c − V L ) 2 = i 2 [ R 2 + ( X c − X L ) 2 ] = 1 5 2 + ( 33.68 − 30.16 ) 2 117 = 7.59 A
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