A steady current I enters a circuit at node P and leaves at node Q . The circuit consists of two parallel branches between P and Q . The upper branch contains a single resistor of resistance R₁ . The lower branch contains two resistors in series, having resistances R₂ and R₃ . If I₂ is the current flowing through R₂ , what is the ratio I₂ I ?

Step-by-step solution

The circuit consists of two parallel branches between nodes P and Q . The resistance of the upper branch is R_ upper = R₁ . The resistance of the lower branch is R_ lower = R₂ + R₃ , as the two resistors are in series. Since the branches are in parallel, the potential difference V across both branches is the same. The current through the lower branch is I₂ = V R₂ + R₃ . The total current I is given by I = V R_ eq , where R_ eq is the equivalent resistance of the circuit. 1 R_ eq = 1 R₁ + 1 R₂ + R₃ = R₁ + R₂ + R₃ R₁