AP EAMCET · PHYSICS · Current Electricity
If \(\mathrm{E}_1=4 \mathrm{~V}\) and \(\mathrm{E}_2=12 \mathrm{~V}\), the current in the circuit and potential difference between the points P and Q respectively are
- A \(1 \mathrm{~A}, 8 \mathrm{~V}\)
- B \(1 \mathrm{~A}, 6 \mathrm{~V}\)
- C \(0.8 \mathrm{~A}, 6.4 \mathrm{~V}\)
- D \(0.8 \mathrm{~A}, 8 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(0.8 \mathrm{~A}, 6.4 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The current in the circuit is \(\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{12-4}{8+1+1}=\frac{8}{10}=0.8 \mathrm{~A}\) \(\therefore \quad\) The potential difference between points \(P\) and \(Q\) is \(V_{P Q}=I R=0.8 \times 8=6.4 \mathrm{~V}\)
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