Consider the following equivalent circuit

Since there is violation of the KVL in the network, the circuit connection is physically not possible.
Applying KVL, we get

5 – 10 = 0 ⇒ −5 = 0

which is not possible.

Now consider the following equivalent circuit

Here applying KVL, we get
10 + 10 = 0 ⇒ 20 = 0
It is also violation of KVL, so physically not possible.
The value of current can be determined using the following circuit.

The voltage at 5 Ω resistor is V_{5  \Omega}=10  V , therefore current is

I_{5  \Omega}=\frac{10}{5}=2  A