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JEE Main 202227 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
27 July
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2022 · 27 July · Morning Shift · Physics · Current Electricity

Two sources of equal emfs are connected in series. This combination...

Two sources of equal emfs are connected in series. This combination is connected to an external resistance RR. The internal resistances of the two sources are r1{r}_{1} and r2(r1>r2){r}_{2}\left({r}_{1} > {r}_{2}\right). If the potential difference across the source of internal resistance r1{r}_{1} is zero then the value of RR will beTwo sources of equal emfs are connected in series. This combination is connected to an external resistance R . The

Options

  1. Ar1−r2{r}_{1} - {r}_{2}
  2. Br1r2r1+r2\dfrac{{r}_{1}{r}_{2}}{{r}_{1} + {r}_{2}}
  3. Cr1+r22\dfrac{{r}_{1} + {r}_{2}}{2}
  4. Dr2−r1{r}_{2} - {r}_{1}

Correct answer

A. r1−r2{r}_{1} - {r}_{2}

Step-by-step solution

Two sources of equal emfs are connected in series. This combination is connected to an external resistance R . The

Current through the wire will be,

I=2εR+r1+r2I = \dfrac{2ε}{R + {r}_{1} + {r}_{2}}.

Now potential difference across cell 11 will be, ε−Ir1=0ε - I{r}_{1} = 0

⇒ε=2εR+r1+r2×r1⇒R+r1+r2=2r1⇒R=r1−r2⇒ ε = \dfrac{2ε}{R + {r}_{1} + {r}_{2}} \times {r}_{1} ⇒ R + {r}_{1} + {r}_{2} = 2{r}_{1} ⇒ R = {r}_{1} - {r}_{2}

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