Skip to content
NCERT Exemplar · Q19

Q.Two cells of the same emf EE but different internal resistances r1r_1 and r2r_2 are connected in series with each other and with an external resistor RR, the first cell, the second cell and RR forming a single series loop. What should be the value of RR so that the potential difference across the terminals of the first cell (the one with internal resistance r1r_1) becomes zero?

Punjab PsebSubjective· 3mImportance★★★★★
83% · 35/42 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The two identical-emf cells in series give a total emf 2E2E driving a current I=2ER+r1+r2I = \dfrac{2E}{R + r_1 + r_2}. The first cell's terminal voltage is E−Ir1E - I r_1; forcing this to zero fixes the current, and solving gives R=r1−r2R = r_1 - r_2.

Concept

For a cell of emf EE and internal resistance r1r_1 carrying (delivering) current II, its terminal potential difference is V=E−Ir1V = E - I r_1. This drops to zero when the internal drop Ir1I r_1 exactly equals the emf.

Step 1 — series current

Total emf =E+E=2E= E + E = 2E; total resistance =R+r1+r2= R + r_1 + r_2. Hence

I=2ER+r1+r2.I = \frac{2E}{R + r_1 + r_2}.

Step 2 — zero terminal voltage of the first cell

V1=E−Ir1=0⟹I=Er1.V_1 = E - I r_1 = 0 \quad\Longrightarrow\quad I = \frac{E}{r_1}.

Step 3 — solve for R

Equate the two expressions for II: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.