A battery of e.m.f. 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.

Find:

(i) Current in the circuit.

(ii) The terminal voltage of the cell.

(iii) The potential difference across 6 Ω Resistor.

(iv) Electrical energy spent per minute in 4 Ω Resistor.

Given,

R_A = 4 Ω

R_B = 6 Ω

e.m.f. = 12 V

r = 2 Ω

(i) Two resistors ( 4 Ω and 6 Ω) are connected in series, so R_s = 4 + 6 = 10 Ω

From relation,

I = $\dfrac{𝐸}{𝑅 + 𝑟}$

Substituting the values we get,

$I = \dfrac{12}{10 + 2} \\[0.5em] I = \dfrac{12}{12} \\[0.5em] \Rightarrow I = 1 A$

(ii) Terminal voltage of cell V = ε - Ir

Substituting the values we get,

$V = 12 - (1 \times 2) \\[0.5em] V = 12 - 2 \\[0.5em] \Rightarrow V = 10 V$

(iii) P.d. across 6 Ω resistor = I R_6Ω

Substituting the values we get,

V_6Ω = 1 x 6 = 6 V

(iv) Electrical energy spent per minute (60 s) in 4 Ω Resistor = I²R_At = 1 x 6 = 6 V