An alternating voltage of 141.4,V (rms) is ap | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The circuit shown is a half-wave rectifier circuit with a capacitor filter.In a half-wave rectifier,the capacitor charges to the peak value of the

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The circuit shown is a half-wave rectifier circuit with a capacitor filter.In a half-wave rectifier,the capacitor charges to the peak value of the input alternating voltage.The relationship between the peak voltage $(V_{peak})$ and the root-mean-square voltage $(V_{rms})$ is given by:$V_{peak} = \sqrt{2} 	imes V_{rms}$Given that $V_{rms} = 141.4\,V$,we calculate the peak voltage as:$V_{peak} = \sqrt{2} 	imes 141.4$Since $\sqrt{2} \approx 1.414$,we have:$V_{peak} = 1.414 	imes 141.4 = 200\,V$Therefore,the maximum potential difference across the condenser is $200\,V$.

An alternating voltage of $141.4\,V$ (rms) is applied to a vacuum diode as shown in the figure. The maximum potential difference across the condenser will be......$V$

Similar Questions

Normally, a capacitor is connected across the output terminals of a rectifier to

The output current versus time curve of a rectifier is shown in the figure. The average value of the output current in this case is

$A$ $220 \,V$ $AC$ supply is connected between points $A$ and $B$ as shown in the figure. What will be the potential difference $V$ across the capacitor?

The value of the form factor in the case of a half-wave rectifier is

For a half-wave rectifier, the load resistance is $R_L = 2 \, k\Omega$ and the $P-N$ junction diode resistance is $R_d = 2 \, k\Omega$. The rectification efficiency is .... $\%$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module