FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Sunday, September 22, 2019

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.


Question 30
(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]





Reference: Past Exam Paper – June 2019 Paper 22 Q5





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

No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | 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.