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Attempt either subpart (a) or (b) :
OR
(b) (i) State the relationship of the resistance (R) of a wire with its area of cross-section (A) and its length (l). With the help of this relation, define resistivity of the material of the wire.
(ii) In an electric circuit, $\text{2 A}$ electric current flows through a $\text{40 \Omega}$ resistor for $\text{50}$ seconds. (I) Calculate the heat produced in the wire. (II) If another $\text{40 \Omega}$ resistor is connected in parallel to the given $\text{40 \Omega}$ resistor, calculate the heat produced in the circuit in $\text{50}$ seconds.
OR
(b) (i) State the relationship of the resistance (R) of a wire with its area of cross-section (A) and its length (l). With the help of this relation, define resistivity of the material of the wire.
(ii) In an electric circuit, $\text{2 A}$ electric current flows through a $\text{40 \Omega}$ resistor for $\text{50}$ seconds. (I) Calculate the heat produced in the wire. (II) If another $\text{40 \Omega}$ resistor is connected in parallel to the given $\text{40 \Omega}$ resistor, calculate the heat produced in the circuit in $\text{50}$ seconds.
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(b) (i) • $R = \rho \frac{l}{A}$ (1 Mark)
• Resistivity of a conductor is the resistance offered by a conductor of length 1 m and area of cross section 1 m² (1 Mark)
(ii) H = $I^2Rt$ (½ Mark)
H = $(2)^2 \times 40 \times 50$ (½ Mark)
H = $8000$ J (½ Mark)
Voltage applied across $40\Omega$, $V=I \times R$ (½ Mark)
= $2 \times 40 = 80$ V (½ Mark)
Equivalent resistance $\frac{1}{R} = \frac{1}{40} + \frac{1}{40}$ (½ Mark)
$R = 20\Omega$ (½ Mark)
Current flowing through circuit
$I = \frac{V}{R}$ (½ Mark)
$I = \frac{80}{20}$ (½ Mark)
$I = 4$ A (½ Mark)
H = $I^2Rt$ (½ Mark)
H = $(4)^2 \times 20 \times 50$ (½ Mark)
H = $16000$ J (½ Mark)
• Resistivity of a conductor is the resistance offered by a conductor of length 1 m and area of cross section 1 m² (1 Mark)
(ii) H = $I^2Rt$ (½ Mark)
H = $(2)^2 \times 40 \times 50$ (½ Mark)
H = $8000$ J (½ Mark)
Voltage applied across $40\Omega$, $V=I \times R$ (½ Mark)
= $2 \times 40 = 80$ V (½ Mark)
Equivalent resistance $\frac{1}{R} = \frac{1}{40} + \frac{1}{40}$ (½ Mark)
$R = 20\Omega$ (½ Mark)
Current flowing through circuit
$I = \frac{V}{R}$ (½ Mark)
$I = \frac{80}{20}$ (½ Mark)
$I = 4$ A (½ Mark)
H = $I^2Rt$ (½ Mark)
H = $(4)^2 \times 20 \times 50$ (½ Mark)
H = $16000$ J (½ Mark)