KCET · Physics · Waves and Sound
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 \( 6: 3: 2 \)
- B \( 6: 4: 3 \)
- C \( 5: 4: 3 \)
- D \( 4: 3: 2 \)
Answer & Solution
Correct Answer
(B) \( 6: 4: 3 \)
Step-by-step Solution
Detailed explanation
According to Ohm's law, we have \( V=\mathbb{R} \).
Since the resistors are connected in parallel, therefore, \( V \) is constant. So,
\(I R=\text { constant } \Rightarrow I \propto \frac{1}{R} \)
\(\Rightarrow I_{1} \propto \frac{1}{2 \Omega}, I_{2} \propto \frac{1}{3 \Omega^{\prime}} I_{3} \propto \frac{1}{4 \Omega} \)
\(\Rightarrow I_{1}: I_{2}: I_{3}=6: 4: 3\)
Since the resistors are connected in parallel, therefore, \( V \) is constant. So,
\(I R=\text { constant } \Rightarrow I \propto \frac{1}{R} \)
\(\Rightarrow I_{1} \propto \frac{1}{2 \Omega}, I_{2} \propto \frac{1}{3 \Omega^{\prime}} I_{3} \propto \frac{1}{4 \Omega} \)
\(\Rightarrow I_{1}: I_{2}: I_{3}=6: 4: 3\)
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