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1 solution
good one sir @Chew-Seong Cheong
But how did you find voltage across R as u already subtracted the voltage then why did you again did it Would u please clarify Mr. @Chew-Seong Cheong
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After deducting the 2V source, it is replaced by a short circuit. The equivalent circuit has only a 4V source at the place of the 6V source. The voltage across R of 1.5V means that on the equivalent circuit.
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Oh thanks sir And would you please tell me this problem W h a t i s t h e l a r g e s t i n t e g e r v a l u e o f n l e s s t h a n 1 0 0 0 s u c h t h a t n = b a − 1 b 2 + a 2 N o t e : a , b a r e n o n n e g a t i v e
please tell me the answer and solution sir @Chew-Seong Cheong
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@Mehul Chaturvedi – I didn't solve this problem.
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@Chew-Seong Cheong – why its easy
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@Mehul Chaturvedi – I didn't think I could solve it and I didn't want to lose points. The question has been posted by Calvin Lin. You shouldn't post it again.
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@Chew-Seong Cheong – Yeas i've deleted it sir and got the answer as 5
Since the 6 V and 2 V voltage sources are in opposite directions, the resultant equivalent voltage source is 6 V − 2 V = 4 V . Since V X Y = 2 . 5 V Then the voltage across R , V R = 4 V − 2 . 5 V = 1 . 5 V . Then we have:
2 0 + 5 R = 2 . 5 1 . 5 ⇒ R = 5 3 × 2 5 = 1 5 Ω