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.
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Given,
RA = 4 Ω
RB = 6 Ω
e.m.f. = 12 V
r = 2 Ω
(i) Two resistors ( 4 Ω and 6 Ω) are connected in series, so Rs = 4 + 6 = 10 Ω
From relation,
I = $\frac{𝐸}{𝑅 + 𝑟}$
Substituting the values we get,
$$\begin{aligned}I = \frac{12}{10 + 2} \\[0.5em] I = \frac{12}{12} \\[0.5em] \Rightarrow I = 1 A\end{aligned}$$
(ii) Terminal voltage of cell V = ε - Ir
Substituting the values we get,
$$\begin{aligned}V = 12 - (1 \times 2) \\[0.5em] V = 12 - 2 \\[0.5em] \Rightarrow V = 10 V\end{aligned}$$
(iii) P.d. across 6 Ω resistor = I R6Ω
Substituting the values we get,
V6Ω = 1 x 6 = 6 V
(iv) Electrical energy spent per minute (60 s) in 4 Ω Resistor = I2RAt = 1 x 4 x 60 = 240 J