Verified Solution • 2025–26 Session moderate

QJEEM21Phe422: JEE Main 2021 Physics - Class 12 JEEM21Phe422 Solution: Two cells of emf 2E and E with internal resistance r 1 and r 2…

Two cells of emf 2E and E with internal resistance r1 and r2 respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

JEE Main 2021 (Online) 17th March Evening Shift Physics - Current Electricity Question 192 English

Multiple Choice Options

  • A r1 −- r2
  • B r12−r2{{{r_1}} \over 2} - {r_2}
  • C r12+r2{{{r_1}} \over 2} + {r_2}
  • D r1 + r2

Step-by-Step Verified Solution

JEE Main 2021 (Online) 17th March Evening Shift Physics - Current Electricity Question 192 English Explanation 1

l=3ER+r1+r2l = {{3E} \over {R + {r_1} + {r_2}}}

JEE Main 2021 (Online) 17th March Evening Shift Physics - Current Electricity Question 192 English Explanation 2

VA=VB{V_A} = {V_B}

2E=i r12E = i\,{r_1}

2E=3ER+r1+r2r12E = {{3E} \over {R + {r_1} + {r_2}}}{r_1}

2R+2r1+2r2=3r12R + 2{r_1} + 2{r_2} = 3{r_1}

R=r12−r2R = {{{r_1}} \over 2} - {r_2}

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