Adapter-2: E2=19V I2=3.42A I2 in AC= 1.5A;

Adapter-1: E1=18.5V I1=3.5A I1 in AC= 1.6A;

E=Voltage; I=Amps; W=Watts;

```
W DC=E DC * I DC;
```

Adapter-2 capacity of electricity:

```
W2=E2*I2=19*3.42=64.98 Watt
```

Adapter-1 capacity of electricity:

```
W1=E1*I1=18.5*3.5=64.75 Watt
```

Deviation accuracy capacity of electricity:

```
(W2/W1)*100%-100%=+0,36%
```

Replacement should be of the same polarity, the difference in the *supply voltage does not exceed 0.5V*, and have sufficient capacity.

# Not use 10% criteria to laptop power supplay!

In general, post it became clear that the power is almost identical ... it was possible to finish the ode Electrical and arithmetic, but we have a little torment to useful conclusions.

Deviation accuracy voltage:

```
(E2/E1)*100%-100%=(19/18.5)*100-100=+2,7%
```

Deviation accuracy DC:

```
(I2/I1)*100%-100%=(3,42/3,5)*100-100=-2,3%
```

The efficiency of conversion from AC to DC:

W1~W2; E1 AC = E2 AC;

```
(I2 in AC/I1 in AC)*100%-100% = (1.5/1.6)*100%-100% = -6,25%
```

Wow! Adapter-2 - very good!!!

Use it. Good luck!

Watch for temperature. Watch for characteristics of supply power at different CPU load as possible.