Find the Vrms across each of the three elements when a 10 ohms resistor, 10mH inductor and 10nF capacitor are connected in series with a 10MHz voltage source. The rms current through the circuit is 0.50A.
The total impedance in the series RLC circuit is
Z=R2+(ωL−1ωC)2=100+(107⋅10−2−1107⋅10−8)2=100+(105−10)2=105 Ω\displaystyle Z = \sqrt{R^2 + (\omega L -\frac{1}{\omega C})^2} = \sqrt{100 + (10^7 \cdot10 ^{-2} -\frac{1}{10^{7} \cdot 10^{-8}})^2} = \sqrt{100+ (10^5-10)^2} = 10^5\; \OmegaZ=R2+(ωL−ωC1)2=100+(107⋅10−2−107⋅10−81)2=100+(105−10)2=105Ω
Then
Vrms=IrmsZ=0.5A105Ω=5⋅10−6 V=5 μV\displaystyle V_{rms} = \frac{I_{rms}}{Z} = \frac{0.5 A}{10^5 \Omega} = 5 \cdot 10^{-6} \; V = 5 \; \mu VVrms=ZIrms=105Ω0.5A=5⋅10−6V=5μV
Answer: 5 μV5 \; \mu V5μV
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