Three resistances \( 2 \Omega, 3 \Omega \) and \( 4 \Omega \) are connected in parallel. The ratio of currents passing
through them when a potential difference is applied across its ends will be

• A

  \( 6: 3: 2 \)
• Correct answer
  B

  \( 6: 4: 3 \)
• C

  \( 5: 4: 3 \)
• D

  \( 4: 3: 2 \)

Solution

According to Ohm's law, we have \( V=\mathbb{R} \).

Since the resistors are connected in parallel, therefore, \( V \) is constant. So,
\[ \begin{array}{l} I R=\text { constant } \Rightarrow I \propto \frac{1}{R} \\ \Rightarrow I_{1} \propto \frac{1}{2 \Omega}, I_{2} \propto \frac{1}{3 \Omega^{\prime}} I_{3} \propto \frac{1}{4 \Omega} \\ \Rightarrow I_{1}: I_{2}: I_{3}=6: 4: 3 \end{array} \]

Hence, the correct answer is option: 'B'.