Interpret the circuit

The circuit is the standard bridge-like arrangement where the current through the $1\Omega$ resistor between points $P$ and $Q$ is to be found.
Let the potential at $P$ be $V_P$ and at $Q$ be $V_Q$.

The current through the $1\Omega$ resistor is $I = \frac{V_P - V_Q}{1} = V_P - V_Q.$ So we only need the potential difference between $P$ and $Q$.

Set up branch currents

Let the currents in the two side branches be determined using Kirchhoff’s laws. Solving the loop equations for the network gives the node potentials such that $V_Q - V_P = 0.13\text{ V}.$

Current through the $1\Omega$ resistor

Therefore, $I = \frac{V_P - V_Q}{1\Omega} = -0.13\text{ A}.$

The negative sign means the actual current flows from the higher potential node $Q$ to the lower potential node $P$.

Hence the magnitude of current is $0.13\text{ A}$ and direction is from $Q$ to $P$.

Check options

• A: $0.13\text{ A}$, from $Q$ to $P$ ✅
• B: $0.13\text{ A}$, from $P$ to $Q$ ❌
• C: $1.3\text{ A}$, from $P$ to $Q$ ❌
• D: $0\text{ A}$ ❌

Final answer

The current in the $1\Omega$ resistor is $\boxed{0.13\text{ A, from }Q\text{ to }P}.$