Question

The circuit diagram shows four dry cells with an e.m.f. 1.5 V and internal resistance 0.25 Ω connected to an external circuit. A 3 Ω wire is immersed in 20 g of water at 20°C. The current switched on for 6 minutes and 36 seconds. Calculate:

1. Reading shown by the ammeter
2. Current in the 1.5 Ω wire
3. Final temperature of water

Answer 1

e.m.f. of each cell = 1.5 V; internal resistance of each cell = 0.25 Ω; mass of water = 20 g

Initial temp = 20°C; time (t) = 6 min 36 s = 396 s

Let θ_R be the rise in temperature of the water.

Total e.m.f. of four cells in series

E = 4 × 1.5 = 6 V

Total internal resistance of four cells

r = 4 × 0.25 = 1 Ω

(a) Resistance of 1.5 Ω and 3 Ω in parallel

\[\frac{1}{R_p} = \frac{1}{1.5} + \frac{1}{3}\]

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

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

\[\therefore R_p = 1\Omega\]

∴ Total external resistance

R = 1 Ω + 3 Ω = 4 Ω

∴ Reading shown by the ammeter

\[I = \frac{E}{R + r}\]

\[= \frac{6}{4 + 1}\]

= 1.2 A

(b) p.d. across the parallel resistors

V = IR_p

= 1.2 × 1

= 1.2 V

∴ Current in the 1.5 Ω wire

\[I = \frac{V}{R}\]

\[= \frac{1.2}{1.5}\]

= 0.8 A

(c) Energy flowing through the 3 Ω wire

\[= I^2Rt\]

\[= (1.2)^2 \times 3 \times 396\]

= 1710.72 J

Energy absorbed by the water

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

\[= 20 \times 4.2 \times \theta_R\]

By the principle of calorimetry

Energy absorbed by the water = Energy flowing through the wire

\[20 \times 4.2 \times \theta_R = 1710.72\]

\[\therefore \theta_R = \frac{1710.72}{20 \times 4.2}\]

\[\therefore \theta_R = 20.36\text{°C}\]

∴ Final temperature of water = Initial temp. + Rise in temp.

\[= 20\text{°C} + 20.36\text{°C}\]

\[= 40.36\text{°C}\]