The resistors are of resistance 15 Ω and 45 Ω. The battery is found to provide 1.6 × 105 J of electrical energy

Question 5

A student set up the circuit shown in Fig. 7.1.

Fig. 7.1

The resistors are of resistance 15 Ω and 45 Ω. The battery is found to provide 1.6 × 10⁵ J of electrical energy when a charge of 1.8 × 10⁴ C passes through the ammeter in a time of 1.3 × 10⁵ s.

(a) Determine

(i) the electromotive force (e.m.f.) of the battery,

(ii) the average current in the circuit.

 [4]

(b) During the time for which the charge is moving, 1.1 × 10⁵ J of energy is dissipated in the 45 Ω resistor.

(i) Determine the energy dissipated in the 15 Ω resistor during the same time.

(ii) Suggest why the total energy provided is greater than that dissipated in the two

resistors.

[4]

Reference: Past Exam Paper – November 2002 Paper 2Q7

Solution:

(a)

(i)

{e.m.f. is the energy converted per unit charge.}

e.m.f. = energy / charge        

e.m.f. = 1.6×10⁵ / 1.8×10⁴

e.m.f. = 8.9 V                         

(ii)

Current I = Q / t                      

Current I = 1.8×10⁴ / 1.3×10⁵

Current I = 0.14 A      

(b)

(i)

{V = IR            and from the definition of e.m.f., V = W / Q

So, W / Q = IR

Since the resistors are in series, the same current flow through them.

For constant current I,

Energy W is proportional to the resistance R.}

Energy ∝ R     (of formula)    

45 Ω - - > 1.1 × 10⁵ J

1 Ω - - - > 1.1 × 10⁵ / 45

15 Ω - - -> (15 / 45) × 1.1 × 10⁵

Energy = 3.7×10⁴ J               

(ii)

Some energy are dissipated in the internal resistance (of the battery)