(a) State Kirchhoff’s second law. (b) A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to

Q#30 (Past Exam Paper – June 2019 Paper 22 Q5)

(a) State Kirchhoff’s second law. [2]

(b) A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to two external resistors, as shown in Fig. 5.1.

Fig. 5.1

The reading on the voltmeter is 4.8 V.

(i) Calculate:

1. the combined resistance of the two resistors connected in parallel [2]

2. the current in the battery. [2]

(ii) Show that the internal resistance r is 2.5 Ω. [2]

(iii) Determine the ratio

power dissipated by internal resistance r

total power produced by battery .

 [3]

(c) The battery in (b) is now connected to a battery of e.m.f. 7.2 V and internal resistance 3.5 Ω.

The new circuit is shown in Fig. 5.2.

Fig. 5.2

Determine the current in the circuit. [2]

 [Total: 13]

Solution:

(a) Kirchhoff’s second law states that the sum of e.m.f.(s) around a loop (or around a closed circuit) is equal to the sum of p.d.(s) in the loop.

(b)

(i)

1.

1 / R = 1 / R1 + 1 / R2

1 / R = 1 / 90 + 1 / 18

R = 15 Ω

2.

I = V / R

{To obtain the current in the battery (this is the total current in the circuit), we need to consider the total e.m.f. (= 5.6 V) in the circuit and the total resistance (= 15 + r). However, we do not have the value of r. So, this method cannot be used.

Instead, consider the terminal p.d. (= 4.8 V) which is equal to the p.d. across the parallel combination of external components (combined resistance = 15 Ω). Alternatively, we could calculate the current in each resistor.}

I = 4.8 / 15                   

or I = 4.8 / 90 + 4.8 / 18

I = 0.32 A

(ii)

{e.m.f. = terminal p.d. V + lost volts (Ir)}

E = V + Ir                    or                    E = I(R + r)

5.6 = 4.8 + 0.32 r        or                     5.6 = 0.32 × (15 + r)

So r = 2.5 Ω

(iii)

P = EI             or P = VI                     or P = I²R                   or P = V² / R

{ratio = I²r / VI}

ratio = (0.32² × 2.5) / (5.6 × 0.32)       or 0.256 / 1.792

ratio = 0.14

(c)

{From Kirchhoff’s second law,

Sum of e.m.f. = sum of p.d.

The positive terminals of the battery are connected to each other, so we must subtract the values.

Let the current in the circuit be I.

7.2 – 5.6 = I × (2.5 + 3.5)}

7.2 – 5.6 – 2.5I – 3.5I = 0

I = 0.27 A

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