Question reads:

A full-wave bridge rectifier with an RL load is connected to a 120V source. If the load resistance is 10.8 \$\Omega\$ and L is very large, find:

(a) Average load voltage

(b) Average load current

(c) Max load current

(d) RMS value of load current

(e) Average current in each diode

(f) RMS current in each diode

(g) Power supplied to the load

(h) Ripple factors of the load voltage and current

(i) Rectifier efficiency


(a) $$V_{o(avg)}=\frac{2V_m}{\pi} =\frac{2\sqrt{2}V_s}{\pi} =108.04 \text{ V} $$

(b) $$\frac{V_{o(avg)}}{R}=\frac{2V_m}{\pi R}=10 \text{ A} $$

(c) $$\text{Should just be when $V_{m}$ is peak and $R$ is minimum, hence, }10 \text{ A} $$

(d) $$\text{Very large $L$ as given in the question}$$ $$\text{therefore the RMS current can be assumed to be equal to $I_{o(avg)}$, hence, }10 \text{ A} $$

(e) $$I_{D(avg)}=\frac{I_{o(avg)}}{2}=5 \text{ A} $$

(f) $$I_{D(RMS)}=\frac{I_{o(avg)}}{\sqrt{2}}=3.54 \text{ A} $$

(g) $$P_o=I_{o(avg)} \cdot V_{o(avg)}=1080.4 \text{ W} $$

(h) $$RF_V=\sqrt{\frac{V^2_{RMS}}{V^2_{o(avg)}}-1} = 0.619$$

$$RF_I=\sqrt{\frac{I^2_{RMS}}{I^2_{o(avg)}}-1} = 0 $$

(i) $$\text{Efficiency}=\frac{V_{o(avg)}\cdot I_{o(avg)}}{V_{RMS}\cdot I_{RMS}} = \frac{108.048\cdot 10}{\frac{120\sqrt{2}}{2}\cdot 10} = 1.273$$

Assuming, \$V_{RMS} = \frac{V_s \sqrt{2}}{2}\$

Obviously the efficiency shouldn't be greater than one... What have I done wrong?

Following Felthry's comment, I notice the above formula for \$V_{RMS}\$ only stands for half-wave rectification, hence, \$V_{RMS} = 120 \text{ V}\$ and the efficiency ~90%.

  • \$\begingroup\$ Right off the bat I can tell you that when something says "a 120V source" without specifying, what is meant is that V_RMS=120V. \$\endgroup\$
    – Hearth
    Apr 30, 2017 at 22:07
  • \$\begingroup\$ @Felthry Ahhh... Note that I need the output RMS voltage, though, not the input RMS voltage. Is it still 120V? I think Vrms = Vs*sqrt(2)/2 only applies to half-wave rectifiers. If so, that was a simple mistake... \$\endgroup\$
    – smollma
    Apr 30, 2017 at 22:57
  • \$\begingroup\$ An ideal full-wave voltage rectifier does not change the RMS voltage. Even a non-ideal one would change it by only a miniscule amount, in this case. \$\endgroup\$
    – Hearth
    Apr 30, 2017 at 22:59
  • \$\begingroup\$ @Felthry Alright, thanks for your help. I'll amend question. \$\endgroup\$
    – smollma
    Apr 30, 2017 at 23:00
  • \$\begingroup\$ I'll make this into an actual answer so that you can mark the question as answered, then. \$\endgroup\$
    – Hearth
    May 1, 2017 at 0:01

1 Answer 1


The formula you use for RMS voltage applies only to a half-wave rectified voltage. The RMS voltage of a full-wave rectified signal is the same as that of the unrectified signal, ignoring diode drops (which are insignificant at 120V).

This can be seen by observing that the first step in calculating the RMS value is squaring the voltage; x2 and (-x)2 are equal for all real values of x, so after this initial squaring, you can't even tell whether the original value was positive or negative. So the RMS value of |V(t)| is necessarily equal to the RMS value of V(t), for any function V.


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