In a circuit, resistance \( R_1 \) is connected in series with the parallel combination of resistances \( R_2 \) and \( R_3 \). The currents passing through the resistances \( R_1 \), \( R_2 \) and \( R_3 \) are \( I_1 \), \( I_2 \) and \( I_3 \), respectively. What will be the ratio \( (I_3/I_1) \) of currents in terms of the resistances connected in the circuit?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( R_2/(R_2 + R_3) \)
The whole current \( I_1 \) passing through \( R_1 \) splits between the two parallel branches, and both branches have the same potential difference.
\( \displaystyle I_1 = I_2 + I_3, \qquad I_2 R_2 = I_3 R_3 \;\Rightarrow\; I_2 = \frac{I_3 R_3}{R_2} \)
\( \displaystyle I_1 = I_3\left(\frac{R_3}{R_2} + 1\right) = I_3\,\frac{R_2 + R_3}{R_2} \)
\( \displaystyle \frac{I_3}{I_1} = \frac{R_2}{R_2 + R_3} \)
This is the current divider rule: a branch takes a share proportional to the resistance of the other branch.
Shortcut: Put \( R_3 \to 0 \). Then \( R_3 \) takes all the current, so \( I_3/I_1 = 1 \). Only \( R_2/(R_2 + R_3) \) gives 1; the other three options do not.