Question

Consider an electric circuit with a diode. What will be the average power dissipated in the resistor? (Assume diode is ideal)
 

 $ \left( A \right)\dfrac{{32{E_0}^2}}{R} \\
  \left( B \right)\dfrac{{16{E_0}^2}}{R} \\
  \left( C \right)\dfrac{{4{E_0}^2}}{R} \\
  \left( D \right)\dfrac{{{E_0}^2}}{{2R}} \\ $

Answer 1

Hint :In order to solve this question, we are going to first consider the two cases for the circuit, one with a diode and the other without a diode, after that, we need to calculate the power of the case without the diode, and then, from the formula , we calculate the power for the circuit with a diode.
The power dissipated by the diode can be calculated as:
 $ {P_D} = \dfrac{1}{2}{P_0} $
Where $ {P_D} $ is the power dissipated with a diode while $ {P_0} $ is the power dissipated without a diode
Power of circuit without diode
  $ {P_0} = \dfrac{1}{2} \times \dfrac{{voltag{e^2}}}{{resistance}} $ .

Complete Step By Step Answer:
Let us start by taking two cases of the circuit, first with a diode which is given in the above circuit and second without a diode.
With a diode:

Without a diode:

The power dissipated by the diode can be calculated as:
 $ {P_D} = \dfrac{1}{2}{P_0} $
Where $ {P_D} $ is the power dissipated with a diode while $ {P_0} $ is the power dissipated without a diode
Now, without the diode, the power is given by:
 $ {P_0} = \dfrac{1}{2} \times \dfrac{{{{\left( {2\dfrac{{{E_0}}}{{\sqrt 2 }}} \right)}^2}}}{{\dfrac{R}{4}}} $
Putting this in the above equation to calculate the power without a diode, we get
 $ {P_D} = \dfrac{1}{2} \times \dfrac{{{{\left( {2\dfrac{{{E_0}}}{{\sqrt 2 }}} \right)}^2}}}{{\dfrac{R}{4}}} = \dfrac{{4{E_0}^2}}{R} $
Hence, option $ \left( C \right)\dfrac{{4{E_0}^2}}{R} $ is the correct answer.

Note :
The key function of an ideal diode is to control the direction of current-flow. Current passing through a diode can only go in one direction, called the forward direction. Current trying to flow the reverse direction is blocked. By applying a diode in the circuit, the power becomes half of that of the power without diode.