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A crystal diode that has an internal resistance of $20 \Omega$ is used tor rectification. It the supply voltage is $50 \sin \omega t$ and the load resistance is $800 \Omega$, then find the rms value of the load current.
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Verified Answer
The correct answer is:
$30.5 \mathrm{~mA}$
Concept:
In a half-wave rectifier,
The average value of load current, $I_{a v g}=\frac{V_m}{\pi\left(R+R_L\right)}$
The RMS value of load current, $I_{r m s}=\frac{V_m}{2\left(R+R_L\right)}$
$V_m$ is the peak value of supply voltage
$R$ is the internal resistance
$R_{\mathrm{L}}$ is the load resistance
Calculation:
Given that, supply voltage $=50 \sin \omega t$
Peal voltage $\left(\mathrm{V}_{\mathrm{m}}\right)=50 \mathrm{~V}$
Internal resistance $(\mathrm{R})=20 \Omega$
Load resistance $\left(R_L\right)=800 \Omega$
The RMS value of load current, $I_{r m s}=\frac{50}{2(20+800)}=30.5 \mathrm{~mA}$
In a half-wave rectifier,
The average value of load current, $I_{a v g}=\frac{V_m}{\pi\left(R+R_L\right)}$
The RMS value of load current, $I_{r m s}=\frac{V_m}{2\left(R+R_L\right)}$
$V_m$ is the peak value of supply voltage
$R$ is the internal resistance
$R_{\mathrm{L}}$ is the load resistance
Calculation:
Given that, supply voltage $=50 \sin \omega t$
Peal voltage $\left(\mathrm{V}_{\mathrm{m}}\right)=50 \mathrm{~V}$
Internal resistance $(\mathrm{R})=20 \Omega$
Load resistance $\left(R_L\right)=800 \Omega$
The RMS value of load current, $I_{r m s}=\frac{50}{2(20+800)}=30.5 \mathrm{~mA}$
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