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
So,
(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%.
