3..

-
 Determine the resistor value required to use 0-1 mA meter with an internal
resistance of 125 Ω for a 0-1 V meter.
Ans. _ e = 1v, I = 1mA, r = 125 Ω
R = e/𝐼 − 𝑟 = 1𝑣 /1𝑚𝐴 − 125 Ω 
= 1000 − 125Ω = 875 Ω 
Hence, the required Resistor value is 875 Ω.

Comments

Post a Comment

Popular posts from this blog

difference between half wave and full wave with respect to filter theory

-
Differentiate between half wave and full wave with respective filter theory?  According to filter theory, the filter convert pulsating direct current into pure current. The working of the full wave  rectifier and half wave rectifier is almost similar with the filter. Since both rectifier use filter to decrease the ripple DC. But the only difference is that in the half rectifier only one half cycle other positive or negative of the input AC current will change the capacitor , but the remaining half cycle will not charge the capacitor.  In the full wave rectifier both positive and negative half cycles of input AC current will charge the capacitor
Read more

Regulates power supply

-
Image
Draw the full regulated power supply and explain its each component.  Unregulated power supply:- In an regulated power supply the regulation of voltage is not reflected in the output since it don't have voltage regulation.    Regulated power supply:- In this regulated circuit a resistor R is connected in series with the input voltage from an and regulated power supply and zener diode connected in parallel with the load.  Components of regulated power supply:- (i) Step down transformer:- It step down down the voltage from AC mains and maintain required level of voltage. The output of Transformer in input of rectifier circuit.  (ii) Rectifier :-  Rectifier is an electronic devices used to convert alternating current coming from Transformer into the corresponding direct current.  (iii)   DC filteration: - DC filtration filters the rippled direct current into pure ripple  free DC waveform.  (iv)   Regulation :-...
Read more

Ripple factor and efficiency

-
Derive the Ripple factor and efficiency in case of half wave and full wave rectifier circuit .  Ripple factor: - Ripple factor is the ratio of RMS value of AC component present in the rectified output to the average value of rectified output. it is  dimensionless quantity.  Rectification efficiency:- it is a quantitative measurement of rectifier efficiency . it is ratio of DC power output to AC power input to the rectifier. it is denoted and expressed in percentage.  Derivation of Ripple factor: By definition, The RMS value of total load current is given by I rms = √( I dc)^2+ ( I ac) ^2 (or) Iac = √ I rms ^2+ I dc^2 When the above equation is divided by using Idc then we can get the following equation. Iac / Idc = 1/ Idc √I2rms + I2dc However, here Iac / Idc is the ripple factor formula R.F = 1/ Idc √I2rms + I2dc = √ (Irms / Idc)2 -1 Ripple Factor of Half Wave Rectifier For half-wave rectifier, Irms = Im/2 Idc = Im/ π We know the formula of R.F = √ (Irms / Idc...
Read more