(i)
The current will decrease to one-third of its initial value.
Ohm's Law: The potential difference (V) across the ends of a metallic conductor is directly proportional to the current (I) flowing through it, provided its temperature remains constant.
$$V \propto I \quad \Rightarrow \quad V = IR \quad \Rightarrow \quad I = \frac{V}{R}$$
Since R is constant and V becomes V/3, the new current $I' = \dfrac{V/3}{R} = \dfrac{I}{3}$.
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(ii)
Circuit diagram: [Battery of 4 cells (4 × 1.5 V = 6 V) — plug key — 5 Ω — 10 Ω — 15 Ω all in series]
Total EMF = 4 × 1.5 = 6 V
Total resistance (series): $R_s = 5 + 10 + 15 = 30\ \Omega$
(I) Current through the circuit:
$$I = \frac{V}{R_s} = \frac{6}{30} = \mathbf{0.2\ A}$$
(II) Potential difference across 10 Ω resistor:
$$V_{10} = I \times R = 0.2 \times 10 = \mathbf{2\ V}$$
Source: Chapter 11 – Electricity, Sections 11.4 (Ohm's Law) and 11.6 (Resistance of a System of Resistors)
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