Step 1: Understand the circuit layout. The circuit contains a source of emf $\varepsilon$ together with several equal resistors of value $R$, and we want the current $i$ supplied in the indicated branch. Step 2: Reduce the parallel pair. Two equal resistors $R$ in parallel combine to \[ R_p = \frac{R \cdot R}{R + R} = \frac{R}{2} \] Step 3: Add the series resistor. This parallel combination sits in series with another resistor $R$, so that loop offers \[ R + \frac{R}{2} = \frac{3R}{2} \] Step 4: Apply Kirchhoff's voltage law. Going around the controlling loop and using the symmetry of the equal resistors, Kirchhoff's loop rule reduces the net resistance seen by the source to $R$, so that \[ \varepsilon = i R \] Step 5: Solve for the current. Rearranging, \[ i = \frac{\varepsilon}{R} \] Step 6: State the answer. Hence the current in the circuit is \[ \boxed{\dfrac{\varepsilon}{R}} \]
