Question:
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2 (r1 > r2). If the potential difference across the source of internal resistance r1 is zero, then the value of R will be :
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2 (r1 > r2). If the potential difference across the source of internal resistance r1 is zero, then the value of R will be :
Updated On: Mar 19, 2025
\(r_1-r_2\)
\(\frac{r_1r_2}{r_1+r_2}\)
\(\frac{r_1+r_2}{2}\)
\(r_2-r_1\)
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The Correct Option is A
Solution and Explanation
The correct answer is (A) : \(r_1-r_2\)
\(ΔV=0 ⇒\frac{2ε}{r_1+r_2+R}r_1=ε\)
\(⇒R=r_1-r_2\)
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Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
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- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
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Applications of Electromagnetic Induction
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- Magnetic Flow Meter