CUET · PHYSICS · PYQ PAPER 2023
An LED has a voltage drop of 2 V across it and a current of 10 mA flows when it operates with a 6 V battery through a limiting resistance R. The value of R is :
- A \(40 k \Omega\)
- B \(4 k \Omega\)
- C \(200 \Omega\)
- D \(400 \Omega\)
Answer & Solution
Correct Answer
(D) \(400 \Omega\)
Step-by-step Solution
Detailed explanation
\(V_R = V_{battery} - V_{LED}\) \(V_R = 6 \text{ V} - 2 \text{ V} = 4 \text{ V}\) \(R = \frac{V_R}{I}\) \(R = \frac{4 \text{ V}}{10 \times 10^{-3} \text{ A}} = 400 \, \Omega\)
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