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
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