Combination of Cells Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Combination of Cells Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-6 | DC Circuits and Measurements. |
| Topics | Numericals on Combination of Cells |
| Academic Session | 2025-2026 |
Numericals on Combination of Cells
Combination of Cells Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements
Que-11. Two cells, each of emf 1.5 V joined in series and send 1.0 A current in an external resistor. If the same cells are joined in parallel, then they send 0.6 A current in the same resistor. What is the internal resistance of each cell?
Ans-11 Let internal resistance be r and external resistance be R
In series combination,
Net emf, E=1.5+1.5 = 3V
Net Resistance = R+2r
E = i(R+2r) ⇒ 3 = 1(R+2r) ⇒ 3 = R+2r ——— (1)
In parallel combination,
Net emf, E = 1.5 V
Net resistance = (R+r/2)
E = i(R+r/2) ⇒ 1.5 = 0.6(R+r/2) ⇒ 5 = 2R+r ———- (2)
Multiply by 2 in eqn (1) we get ,
6 = 2R+4r —— (3)
From equation 3 and 2, we get:-
1 = 3r
r = 1/3
Que-12. Two identical cells, whether joined together in series or in parallel, give the same current when connected to an external resistance of 1 Ω. Find internal resistance of each cell.
Ans-12 In series, current, i1 = 2E/(1+2r)
In parallel, current, i2 = E/(1+r/2) = 2E/(2+r)
According to the question
Since, i1−i2 ⇒ E/(2+r) = E/(1+2r)
⇒ r = 1Ω.
Que-13. 50 bulbs are joined in parallel, each being a 100 V bulb taking a current of 0.6 A. A battery of 54 accumulators, each of emf 2 V and internal resistance 0.005 Ω, is used to light up the bulbs. Will the bulbs be properly lighted, over-lit or under-lit?
Ans-13 total current taken by all bulb
= 50 x 0.6
= 30 amp
again current given by all accumulator
= (54 x 2)/{54 x 0.005 + (100/30)}
= 29.97 = 30A
Bulb will be properly lighted.
Que-14. (i) Three cells of emf’s 2.0 V, 1.8 V and 1.5 V and internal resistances 0.05 Ω, 0.7 Ω, and 1.0 Ω, respectively are joined in series. This battery is connected to an external resistor of 4.0 Ω, through a low-resistance ammeter. What would the ammeter read? (ii) If these cells are joined in parallel, can the battery so formed may give a definite emf and internal resistance
Ans-14 (i) i = {(E1+E2+E3)/(R+r1+r2+r3)}
= {(2+1.8+1.5)/(4+0.05+0.7+1.0)}
= 5.3/5.75 = 0.92 A.
(ii) No, for non-similar cells in parallel, Kirchhoff’s rules are employed to get the currents indifferent branches.
Que-15. Three identical cells, each of emf 2 V and unknown internal resistance are connected in parallel. This combination is connected to a 6 Ω resistor. The terminal potential difference across each cell is 1.5 V. What is the internal resistance of each cell?
Ans-15 E = 2V
R = 6 Ω
P.D. = 1.5 V
V = ER/(r+R)
Terminal voltage,
V = ER/{(r/3)+R}
⇒ 1.5 = (2×6){(r/3)+6}
∴ r = 6 Ω.
Que-16. The plot of the variation of potential difference across a combination of three identical cells in series versus current is shown below. What is the emf and internal resistance of each cell?
Ans-16 Since three identical cells are in series.
Emf resultant = emf1 + emf2 + emf3
6 = 3emf
emf = 2V
R = V/I = 6V/2A = 6Ω
Resultant = (R1+R2+R3)
6 = 6R
R = 1Ω.
— : End of Combination of Cells Numerical Class-12 Nootan ISC Physics Ch-6 DC Circuits and Measurements.. :–
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