Explain why the terminal potential difference (p.d.) of a cell with internal resistance may be less than the electromotive force (e.m.f.) of the cell.

Question 12

(a) Explain why the terminal potential difference (p.d.) of a cell with internal resistance may be less than the electromotive force (e.m.f.) of the cell. [2]

(b) A battery of e.m.f. 4.5 V and internal resistance r is connected in series with a resistor of resistance 6.0 Ω, as shown in Fig. 5.1.

 Fig. 5.1

The current I in the circuit is 0.65 A.

Determine

(i) the internal resistance r of the battery, [2]

(ii) the terminal p.d. of the battery, [2]

(iii) the power dissipated in the resistor, [2]

(iv) the efficiency of the battery. [2]

(c) A second resistor of resistance 20 Ω is connected in parallel with the 6.0 Ω resistor in Fig. 5.1.

Describe and explain qualitatively the change in the heating effect within the battery. [3]

Reference: Past Exam Paper – June 2014 Paper 23 Q5

Solution:

(a)

There are lost volts / energy used within the cell / internal resistance when the cell supplies a current.

(b)

(i)

e.m.f. E = I × (R + r)  

4.5 = 0.65 × (6.0 + r)

Internal resistance, r = 0.92 Ω

(ii)

{The terminal p.d. is the p.d. available for the external circuit (after the lost volts in the internal resistance has been accounted for).

So, terminal p.d. is the p.d. that would flow in the resistor.

Thus, when using the terminal p.d., we do not take the internal resistance into account.}

V = IR             and current I = 0.65 A           

Terminal p.d. = 0.65 × 6 = 3.9 V

(iii)

Power dissipated, P = V²/R    or P = I²R        or P = IV

P = (3.9)² / 6 = 2.5 W

(iv)

Efficiency of battery = power out / power in            

Efficiency of battery = I²R / I²(R + r) = R / (R + r) = 6.0 / (6.0 + 0.92) = 0.87

(c)

{Adding resistors in parallel causes the overall resistance to decrease.}

The (circuit) resistance decreases. So, the current increases, causing more heating effect.