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?