Determine the current through the 5 Ω resistor in the following figure.
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