Question

A battery of 12 V and negligible internal resistance is connected to an external circuit consisting of three resistors of 6 Ω, 3 Ω and 2 Ω in parallel, which further connected to a resistance of 3 Ω in series to the battery. The 3 Ω resistance is immersed in 50 g oil of sp. heat capacity 0.8 Jg⁻¹°C⁻¹, when the temperature of the oil rises by 54°C.

1. Draw the labelled circuit diagram.
2. Calculate the value of the current in the main circuit.
3. Calculate the current flowing in 2 resistance in parallel.
4. Calculate the time for which the current is switched on.

Answer 1

(a)

e.m.f. = 12 V; mass of oil = 50 g; sp. heat capacity of oil = 0.8 Jg⁻¹°C⁻¹

Rise in temperature of oil = 54°C

(b) Resistance of 6 Ω, 3 Ω and 2 Ω in parallel

\[\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} + \frac{1}{2}\]

\[\frac{1}{R_p} = \frac{1 + 2 + 3}{6}\]

\[\frac{1}{R_p} = \frac{6}{6}\]

\[\therefore R_p = 1\Omega\]

∴ Total external resistance

\[R = 1\Omega + 3\Omega = 4\Omega\]

∴ Current flowing through the main circuit

\[I = \frac{E}{R} = \frac{12}{4} = 3\text{A}\]

(c) p.d. across the parallel resistors

\[V = IR_p = 3 \times 1 = 3\text{V}\]

∴ Current in 2 Ω resistor

\[I = \frac{V}{R} = \frac{3}{2} = 1.5\text{A}\]

(d) Energy flowing through the 3 Ω wire

\[= I^2Rt\]

\[= (3)^2 \times 3 \times t\]

\[= 27t\text{J}\]

Energy absorbed by the oil

\[= m\,c\,\theta_R\]

\[= 50 \times 0.8 \times 54\]

\[= 2160\text{J}\]

By the principle of calorimetry

Energy absorbed by the oil = Energy flowing through the wire

\[27t = 2160\]

\[\therefore t = \frac{2160}{27}\]

\[\therefore t = 80\text{s}\]