Question

Consider the full-wave bridge rectifier circuit. The input signal is 120 V (rms) at 60 Hz. The load resistance is RL = 250 . The peak output voltage is to be 9 V and the ripple voltage is to be no more than 5 percent. Determine the required turns ratio and the load current.
Basic electronics question. ​

Answer 1

Answer:

आएआम है और वह पैसा कमाने वाले क्रांतिकारी समर्पित कर्तव्यनिष्ठा जुझारू नेता हैं जो वैश्य और आग्नेय में सम्पूर्ण वैश्य और आग्नेय में सम्पूर्ण विश्व में सबसे ज्यादा जिम्मेदार है केंद्र सरकार

Answer 2

Given:

Given,

$\[{{v}_{o\left( \max \right)}}=9V\]$

$\[{{v}_{I,rms}}=120V(rms)\]$

To find:

We need to find the turns ratio and load current.

Solution:

We know that, $\[{{v}_{s(max)}}={{v}_{0(max)}}+2V\gamma \]$.

Where, $\[V\gamma \]$ is the diode cut in voltage and

$\[{{v}_{o\left( \max \right)}}\]$ is desired peak price of the output voltage.

So,

$& {{v}_{\gamma }}=0.7V \\ & \\ & {{v}_{o\left( \max \right)}}=9V \\ & \\ & {{v}_{s\left( \max \right)}}=9+(2\times 0.7)=10.4V$

Now, we will find the turn ratio.

We know that, $\[k=\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{v}_{I,rms}}}{{{v}_{s,rms}}}\]$

Where, $\[{{v}_{s,rms}}=\frac{{{v}_{o\left( \max \right)}}}{\sqrt{2}}\]$

N1 is the range of winding on the first lacet of the transformer.

N2 is the range of winding on the secondary side of the transformer.

So, $\[k=\frac{120}{7.35}=16.3\]$

Therefore, the required turn’s ratio is $16.3$.