(a) (i) The potential difference across the two ends of a circuit component is decreased to one-third of its initial value, while its resistance remains constant. What change will be observed in the current flowing through it? Name and state the law which helps us to answer this question. (ii) Draw a schematic diagram of a circuit consisting of a battery of four $1.5 \mathrm{~V}$ cells, a $5 \Omega$ resistor, a $10 \Omega$ resistor and a $15 \Omega$ resistor and a plug key, all connected in series. Now find (I) the electric current passing through the circuit, and (II) potential difference across the $10 \Omega$ resistor when the plug key is closed. OR (b) (i) When is the potential difference between two points said to be 1 volt? (ii) A copper wire has a diameter of $0.2 \mathrm{~mm}$ and resistivity of $1.6 \times 10^{-8} \Omega \mathrm{~m}$. What will be the length of this wire to make its resistance $14 \Omega$? How much does the resistance change, if the diameter of the wire is doubled?

(a)(i) By Ohm's law, $V = IR$, so at constant resistance the current is directly proportional to the potential difference. If $V$ is reduced to one-third, the current also falls to one-third of its initial value.

Ohm's law: The current through a conductor is directly proportional to the potential difference across its ends, provided its temperature and physical conditions remain constant, i.e. $V = IR$.

(a)(ii) Four cells in series give $V = 4 \times 1.5\,\text{V} = 6\,\text{V}$. In series the resistances add: $R = 5 + 10 + 15 = 30\,\Omega$.

(I) $I = \dfrac{V}{R} = \dfrac{6}{30} = 0.2\,\text{A}$.

(II) Potential difference across the $10\,\Omega$ resistor: $V_{10} = IR = 0.2 \times 10 = 2\,\text{V}$.

OR

(b)(i) The potential difference between two points is 1 volt when 1 joule of work is done to move a charge of 1 coulomb from one point to the other: $1\,\text{V} = 1\,\text{J C}^{-1}$.

(b)(ii) Diameter $d = 0.2\,\text{mm} = 2\times10^{-4}\,\text{m}$, so radius $r = 1\times10^{-4}\,\text{m}$ and area $A = \pi r^{2} = \pi(10^{-4})^{2} = \pi\times10^{-8}\,\text{m}^{2}$. From $R = \dfrac{\rho L}{A}$, $L = \dfrac{RA}{\rho} = \dfrac{14 \times \pi\times10^{-8}}{1.6\times10^{-8}} \approx 27.5\,\text{m}$.

If the diameter is doubled, the area becomes $4A$, and since $R \propto \dfrac{1}{A}$, the resistance becomes one-fourth: $R' = \dfrac{14}{4} = 3.5\,\Omega$ (a decrease of $10.5\,\Omega$).