Question III.1: (a) Explain how ideal voltage and current sources differ fro...
(a) Explain how ideal voltage and current sources differ from practical voltage and current sources, respectively.
(b) Calculate the range of current flowing through the resistance R when its value is varied from 6 ohms to 36 ohms in the circuit shown in Fig. 23 using Thevenin’s theorem.
(a) (i) Refer to Section 2.2.2 on Pages 2.3–2.4.
(b) The open circuit voltage, V_{OC} across the terminals A and B (Fig. 25) by removing the resistance R is calculated as
I_{1}=\frac{90}{60+30}=1 A, I_{2}=\frac{100}{60+40}=1 AThe voltage drop across the 30 Ω resistor = 30 × 1 = 30 V. The potential of point P with respect to N is + 30 V. The potential of point A with respect of point P is +50 volts. Therefore, the potential of point A with respect to point N is +30 +50 = +80 V.
Voltage drop across the 60 Ω resistor in the right-hand-side loop is calculated as
\begin{aligned}I_{2} &=\frac{100}{40+60}=1 A\\ & V_{NB}=60 \times 1=60 V\end{aligned}The potential of point B with respect to point N is +60 V and that of A with respect to N = 30I_{1} + 50 = +80 V.
Now we observe that the potential of point A with respect to point N is +80 V and the potential of point B with respect to point N is +60 V. Therefore, the potential of point A with respect to point B, i.e. V_{OC} becomes +20 V, point A being at a higher potential than point B.
To calculate R_{eq} we redraw the circuit by short circuiting the sources of EMFs as in Fig. 26.
R_{eq}=\frac{60 \times 30}{60+30}+\frac{60 \times 40}{60+40}=44 \OmegaThevenin’s equivalent circuit is drawn as in Fig. 27.
I=\frac{20}{44+6} when R = 6 Ω
= 0.4 A
I=\frac{20}{44+36} when R = 36
= 0.25 A
The value of current through the resistor R will vary from 0.4 A to 0.25 A when its value is changed from 6 Ω to 36 Ω keeping the other circuit conditions unchanged.
