Current Electricity • NEET Revision

Example 3.38

Cells in Series (Aiding)

Question

In the circuit shown, \(E_1 = 10\ \text{V},\ E_2 = 4\ \text{V},\ r_1 = r_2 = 1\ \Omega\) and \(R = 2\ \Omega\). Find the potential difference across battery 1 and battery 2. (Both batteries are connected with the same polarity — they aid each other.)

E₁r₁E₂r₂R = 2 Ω

Given

Asked

Terminal potential difference \(V_1\) across battery 1 and \(V_2\) across battery 2.

Formula

\(I = \dfrac{E_1 + E_2}{r_1 + r_2 + R}\)
Discharging battery: \(V = E - Ir\)

Solution

  1. Net emf \(= E_1 + E_2 = 10 + 4 = 14\ \text{V}\); total resistance \(= 1 + 1 + 2 = 4\ \Omega\).
  2. Current: \(I = \dfrac{14}{4} = 3.5\ \text{A}\).
  3. Both batteries are discharging (delivering current), so each loses \(Ir\) internally:
  4. \(V_1 = E_1 - I r_1 = 10 - 3.5 \times 1 = 6.5\ \text{V}\).
  5. \(V_2 = E_2 - I r_2 = 4 - 3.5 \times 1 = 0.5\ \text{V}\).
  6. Check: \(V_1 + V_2 = 7\ \text{V} = IR = 3.5 \times 2\ \checkmark\)
\(V_1 = 6.5\ \text{V},\quad V_2 = 0.5\ \text{V}\)

Short Cut

Verification habit: the sum of terminal PDs of the sources must equal the PD across the external resistance (\(IR\)). Here \(6.5 + 0.5 = 7 = 3.5\times 2\) — instant self-check in the exam.