KCET · Physics · Current Electricity
Two heating coils of resistances \( 10 \Omega \) and \( 20 \Omega \) are connected in parallel and connected to a battery of emf \( 12 \mathrm{~V} \) and internal resistance \( 1 \Omega \). The power consumed by them are in the ratio
- A \(1: 4\)
- B \( 1: 3 \)
- C \( 1 \)
- D \( 04: 1 \)
Answer & Solution
Correct Answer
(C) \( 1 \)
Step-by-step Solution
Detailed explanation
Since the resistors are connected in parallel, then potential difference across each resistor is same. Power is given as
\(\begin{array}{l}P=\frac{V^{2}}{R} \\\Rightarrow P \propto \frac{1}{R} \\\Rightarrow \frac{P_{1}}{P_{2}}=\frac{R_{2}}{R_{1}} \\\Rightarrow \frac{P_{1}}{P_{2}}=\frac{20}{10}=\frac{2}{1}\end{array}\)
Therefore, the power consumed by them are in the ratio \( 2: 1 \).
\(\begin{array}{l}P=\frac{V^{2}}{R} \\\Rightarrow P \propto \frac{1}{R} \\\Rightarrow \frac{P_{1}}{P_{2}}=\frac{R_{2}}{R_{1}} \\\Rightarrow \frac{P_{1}}{P_{2}}=\frac{20}{10}=\frac{2}{1}\end{array}\)
Therefore, the power consumed by them are in the ratio \( 2: 1 \).
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