(a) Fig. shows two resistances R₁ and R₂ connected in series with a battery of V volts.
Let the p.d. across R₁ is V₁ and the p.d. across R₂ is V₂
s.t. V = V₁ + V₂ ..........(1)

Let the equivalent resistance be R and current flowing through the whole circuit is I
By Ohm's law,
$\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$
V = I × R ..........(2)
Applying ohm's law to both R₁ and R₂
V₁ = I × R₁ ..........(3)
V₂ = I × R₂ ..........(4)
From eqs. (1), (2), (3) and (4), we get
I × R = I × R₁ + I × R₂
I × R = I × (R₁ + R₂)
R = R₁ + R₂
(b) (i) Current through 5Ω resistor = $\frac{10}{5}=2 \mathrm{A}$
(ii) Since 5Ω resistor and R are connected in series , so same current through them. So, current through R = 2 A
(iii) V = IR