The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20 Ω and 5 Ω, is connected to the parallel combination of two resistors 30 Ω and R Ω. The voltage difference across the battery of internal resistance 20 Ω is zero, the value of R (in Ω) is: _______ Numerical answer type. Work it out, then open the answer.

Answer: 30. P.d. across the 20 Ω battery is zero: 10 - 20I = 0 ⇒ I = 0.5 A. Whole circuit: 20 = I(20 + 5 + Rp) ⇒ 25 + Rp = 40 ⇒ Rp = 15 Ω. (30R)/(30 + R) = 15 ⇒ 30R = 450 + 15R R = 30 Ω

The series combination of two batteries, both of the same emf 10 V… | JEE Main 2020 (8 Jan, Shift 2)

Q30JEE Main 2020 · 8 JAN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

The series combination of two batteries, both of the same emf 10 V, but different internal resistance of 20 $\Omega$ and 5 $\Omega$ , is connected to the parallel combination of two resistors 30 $\Omega$ and R $\Omega$ . The voltage difference across the battery of internal resistance 20 $\Omega$ is zero, the value of R (in $\Omega$ ) is: _______

Numerical answer type. Work it out, then open the answer.

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Answer: 30

P.d. across the $20\ \Omega$ battery is zero: $10 - 20I = 0 \Rightarrow I = 0.5\text{ A}$ .
Whole circuit: $20 = I(20 + 5 + R_p) \Rightarrow 25 + R_p = 40 \Rightarrow R_p = 15\ \Omega$ .
$\dfrac{30R}{30 + R} = 15 \Rightarrow 30R = 450 + 15R$
$R = 30\ \Omega$

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