Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.
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$\text{R}=300\Omega$
$\text{C}=20\mu\text{F}=20\times10^{-6}\text{F}$
$\text{L}=1\text{H},\text{Z}=500$ $(\text{from}\ 14)$
$\in_0=50\text{V},\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$
Electric Energy stored in Capacitor $=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times20\times10^{-6}\times50\times50$
$=25\times10^{-3}\text{J}=25\text{mJ}$
Magnetic field energy stored in the coil $=\Big(\frac{1}{2}\Big)\text{LI}_0{^2}$
$=\Big(\frac{1}{2}\Big)\times1\times(0.1)^{2}$
$=5\times10^{-3}\text{J}=5\text{mJ}$
$\text{C}=20\mu\text{F}=20\times10^{-6}\text{F}$
$\text{L}=1\text{H},\text{Z}=500$ $(\text{from}\ 14)$
$\in_0=50\text{V},\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$
Electric Energy stored in Capacitor $=\Big(\frac{1}{2}\Big)\text{CV}^2$
$=\Big(\frac{1}{2}\Big)\times20\times10^{-6}\times50\times50$
$=25\times10^{-3}\text{J}=25\text{mJ}$
Magnetic field energy stored in the coil $=\Big(\frac{1}{2}\Big)\text{LI}_0{^2}$
$=\Big(\frac{1}{2}\Big)\times1\times(0.1)^{2}$
$=5\times10^{-3}\text{J}=5\text{mJ}$
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