MHT CET · Physics · Alternating Current
The a.c. source is connected to series LCR circuit. If voltage across \(\mathrm{R}\) is \(40 \mathrm{~V}\), that across \(\mathrm{L}\) is \(80 \mathrm{~V}\) and that across \(\mathrm{C}\) is \(40 \mathrm{~V}\), then the e.m.f. 'e' of a.c. source is
- A \(40 \mathrm{~V}\)
- B \(40 \sqrt{2} \mathrm{~V}\)
- C \(80 \mathrm{~V}\)
- D \(160 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(40 \sqrt{2} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The emf across the AC source is given as:
\(\mathrm{e} =\sqrt{\left(\mathrm{V}_{\mathrm{K}}\right)^2+\left(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{L}}\right)^2}\) \(=\sqrt{(40)^2+(80-40)^2}\)
\(\mathrm{e} =\sqrt{3200}\)
\(\therefore \mathrm{e} =40 \sqrt{2} \mathrm{~V}\)
\(\mathrm{e} =\sqrt{\left(\mathrm{V}_{\mathrm{K}}\right)^2+\left(\mathrm{V}_{\mathrm{c}}-\mathrm{V}_{\mathrm{L}}\right)^2}\) \(=\sqrt{(40)^2+(80-40)^2}\)
\(\mathrm{e} =\sqrt{3200}\)
\(\therefore \mathrm{e} =40 \sqrt{2} \mathrm{~V}\)
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