Question

A battery of 24 V and negligible internal resistance is connected to two resistors of 12 Ω and 6 Ω in parallel and then to a nichrome wire of resistance 2 Ω, immersed in 100 g of oil (sp. heat capacity 1.2 Jg⁻¹°C⁻¹) at 30°C. The current is switched on for 2 minutes.

1. Draw the circuit diagram of the arrangement.
2. Calculate the current flowing through the main circuit.
3. Current flowing through the 6 Ω resistor.
4. Final temperature of the oil.

Answer 1

(a) e.m.f. = 24 V; mass of oil = 100 g; sp. heat capacity of oil = 1.2 Jg⁻¹°C⁻¹

Initial temp = 30°C; time (t) = 2 min = 120 s.

Let θ_R be the rise in temperature of oil.

(b) Resistance of 12 Ω and 6 Ω in parallel

\[\frac{1}{R_{p}} =\frac{1}{12}+\frac{1}{6}=\frac{3}{12}=\frac{1}{4}\]

∴ R_p = 4 Ω

∴ Total external resistance = 4 Ω + 2 Ω = 6 Ω

∴ Current flowing through the main circuit

\[ I=\frac{E}{R}=\frac{24}{6}= {4 A}\]

(c) p.d. across the parallel resistors

V = IR = 4 × 4 = 16 V

∴ Current in 6 Ω resistor, \[\text{I} = \frac{\text{V}}{\text{R}}\]

\[= \frac{16}{6}\]

= 2.67 А

(d) Energy flowing through the 2 Ω wire = I².R.t

= (4)² × 2 × 120

= 3840 J

Energy absorbed by the oil = m c θ_R = 100 × 1.2 × θ_R

By the principle of calorimetry

Energy absorbed by the oil = Energy flowing through the wire

100 × 1.2 × θ_R = 3840

\[ \theta_{\text{R}}=\frac{3840}{100 \times 1.2}\]

∴ \[ \theta_{\text{R}}=32^{\circ} \text{C} \]

∴ Final temperature of oil = Initial temp. + Rise in temp. (θ_R)

= 30°C + 32°C

= 62°C