**Circuit Diagram:**

The circuit diagram of the given electric circuit is as follows:

```
_______
| |
____| CELL |____
| |_______| |
| |
|K| |
|e| |A|
|y| |m|
|_| |m|
| |
|______| |______|
|
R1
(4Ω)
|
|______|
|
R2
(8Ω)
|______|
|
R3
(8Ω)
|______|
|
|______|
|
|V|
|o|
|l|
|t|
|_|
```

**Explanation:**

To find the maximum current that can flow through the three resistors, we need to consider the maximum power dissipation and the maximum power that the resistors can withstand without melting.

Given that the maximum power that each resistor can withstand is 16W, we can use the formula for power dissipation in a resistor:

```
Power (P) = (Current)^2 * Resistance
```

Since the maximum power dissipation is 16W, we can rearrange the formula to solve for the maximum current (I):

```
16 = I^2 * Resistance
```

Substituting the resistance values for the three resistors, we get:

For R1 (4Ω):
```
16 = I^2 * 4
```

For R2 (8Ω):
```
16 = I^2 * 8
```

For R3 (8Ω):
```
16 = I^2 * 8
```

Simplifying the equations, we get:

For R1:
```
4 = I^2
```

For R2:
```
2 = I^2
```

For R3:
```
2 = I^2
```

Taking the square root of both sides, we find that the maximum current (I) flowing through the resistors is:

For R1:
```
I = √4
I = 2A
```

For R2:
```
I = √2
I = 1.414A
```

For R3:
```
I = √2
I = 1.414A
```

Therefore, the maximum current that can flow through the three resistors is 2A for R1 and 1.414A for R2 and R3.

Note: The voltmeter is connected across the parallel combination of R2 and R3 to measure the voltage drop across them.