There are a large number of cells available, each marked \) to be used to supply current to a device of resistance requiring current. How should the cells be arranged, so that power is transmitted to the load using minimum number of cells?

Four rows, each containing six cells

Physics · Internal resistance, potential difference and emf of a cell, a combination of cells in series and parallel

There are a large number of cells available, each marked \) to be used to supply

There are a large number of cells available, each marked \( (6 V, 0.5 \Omega) \) to be used to supply current to a device of resistance \( 0.75 \Omega, \) requiring \( 24 A \) current. How should the cells be arranged, so that power is transmitted to the load using minimum number of cells?

A. Six rows, each containing four cells
B. Four rows, each containing six cells
C. Four rows, each containing four cells
D. Six rows, each containing Six cells

Step-by-step solution

Let m be number of rows (parallel) and n be cells per row (series). Each cell: emf=6 V, internal resistance=0.5 Ω. Load resistance R_L=0.75 Ω. Equivalent emf = n*6 V, internal resistance = (n*0.5)/m. Current I = (6n) / (0.5n/m + 0.75). Set I=24 A. Simplify to (n-3)(m-2)=6. Integer solutions: (n,m) = (4,8),(5,5),(6,4),(9,3). Total cells N=n*m: 32,25,24,27. Minimum N=24 with n=6,m=4 (4 rows of 6 cells). Among options, B matches. A gives 6×4=24 cells but I≈22.15 A (<24), C gives 4×4=16 cells, I=19.2 A, D gives 6×6=36 cells, I=28.8 A. Thus B is correct.

Practise more in this unit View MCQs Sign up for full question bank

Step-by-step solution

Related MCQs