(a) Two cells of emf E₁ and E₂ have their internal resistances r₁ and r₂ , respectively.
Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance R. Assume that the two cells are supporting each other.
(b) In case the two cells are identical, each of emf E = 5 V and internal resistance r = 2 $Ω$ , calculate the voltage across the external resistance R = 10 $Ω$ .

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Current drawn from cell -1
𝐼1 = (𝐸1 − 𝑉) / 𝑟1
Current drawn from cell -2
𝐼2 = (𝐸2 – 𝑉) / 𝑟2
Resultant current 𝐼 = 𝐼1 + 𝐼2
On solving
$\begin{array}{l} ∴ I = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}} - V (\frac{r_{2} + r_{1}}{r_{1} r_{2}}) \\ ∴ V = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} r_{2}} - I (\frac{r_{1} r_{2}}{r_{2} + r_{1}}) \\ V = E_{eq} - {Ir}_{eq} \\ E_{eq} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} \\ r_{eq} = \frac{r_{1} r_{2}}{r_{2} + r_{1}} \end{array}$
Current through R $r_{eff} = \frac{r_{1} r_{2}}{r_{1} + r_{2}} = \frac{2 \times 2}{2 + 2} = 1 Ω$
P.D across R $= \frac{5}{11} \times 10 = 4 .54 volt$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th