Question

A battery of 4 cell, each of e.m.f. 1.5 volt and internal resistance 0.5 Ω is connected to three resistances as shown in the figure. Calculate:
(i) The total resistance of the circuit.
(ii) The current through the cell.
(iii) The current through each resistance.
(iv) The p.d. across each resistance.

Answer 1

Let total e.m.f. of 4 cells = nE (n = number of cells)

E = 4 × 1.5

E = 6 volts    ....(i)

Total internal resistance = nr (n = 4, r = 0.5 Ω)

Total internal resistance = 4 × 0.5 (in series)

= 2 Ω                    ....(ii)

Let total external resistance = X Ω

`1/"R" = 1/"R"_1 + 1/"R"_2` (R₁ and R₂ are in parallel)

`1/"R" = 1/4 + 1/12 = (3 + 1)/12 = 4/12`

R = `12/4 = 3 Ω`

Total external resistance = X = (R + R₃)Ω (in series)

X = 3Ω + 7 Ω = 10Ω       .....(iii)

Total resistance of the circuit = X + r = 10 + 2 = 12 Ω

Total current through the cell = `"E"/("X" + "r")`   ...(as E = I (R + r))

I = `6/(10 + 2) = 0.5` A      ....from (i), (ii) and (iii)

Current through resistor 4 Ω =

I₁ = `("I" xx "R"_2)/("R"_1 + "R"_2)`

I₁ = `(0.5 xx 12)/(4 + 12) = (0.5 xx 12)/16 = (0.5 xx 3)/4`

I₁ = `1.5 xx 1/4 = 0.375`A

Similarly current through resistor 12 Ω = I₂

I₂ = `"I" xx "R"_1/("R"_1 + "R"_2)`

I₂ = `0.5 xx 4/16 = 0.5 xx 1/4 = 1/8`

I₂ = 0.125 A.

P.D. across resistance 7 Ω = V = I × R = 0.5 × 7 = 3.5 V

P.D. across resistance 4 Ω = V₁ = V₁ × R₁ = 0.375 × 4 = 1.5 V

P.D. across resistance 12 Ω = V₂ = I₂ × R₂ = 0.125 × 12 = 1.5 V.