Thevenin’s theorem is quite useful when the current in one branch of a network is to be determined or when the current in an added branch is to be calculated.
“It states that for the purpose of determining the current in a resistor, RL connected across two terminals of a network which contains sources of e.m.f. and resistors, the network can be replaced by a single source of e. m.f. and a series resistor, R th· This e. m.f., Eth is equal to potential difference between the terminals of the network when the resistor, R, is removed: the resistance of series resistor, R th, is equal to the equivalent resistance of the network with the resistor,· R, removed (or as it is sometimes called, “the resistance of the network when viewed from the terminals under consideration”)”.
Hence I= E /(RL + R th)
Explanation. Let us consider the circuit shown in Fig. 66 (a). The following steps are required to find current through the load resistance RL.
1. Remove RL from the circuit terminals A and B and redraw the circuit as shown in Fig. 66 (b). Obviously the terminals have been open circuited.
2. Calculate the open-circuit voltage (V oc = Eth ) which appears across terminals A and B, when they are open i.e. when RL is removed. This voltage is Eth (Thevenin’s voltage). A little thought will reveal that
3. Short circuit the battery and find the Thevenin resistance R th of the network as seen from the terminals A and B [Fig. 66 (c))
R th = R1 x R2 / R1 + R2
4. Connect RL back across the terminals A and B. from where it was temporarily removed earlier [Fig. 66 (d)J. Current through RL is given by
I = E th / ( R th + RL)
Example 18. With reference to the network shown in Fig. 67, by using Thevenin’s theorem find the following :
(i) The equivalent e.m.f. of the network when viewed from terminals Land M.
(ii) The equivalent resistance of the network when looked into from terminals L and M.
(iii) Current in the load resistance RL of 30 n.
Solution. (i) Equivalent e.m.f. of the network :
Refer Fig. 67.
Current in the network before load resistance (RL) is connected
= 48 / 24 + 6 + 2 = 1.5 A
Voltage across terminals LM, V oc = Eth = 24 x 1.5 = 38 V.
Hence, so far as terminals L and Mare connected, the network has an e.m.f. of 38 V (and not 48 V). (Ans.)
(ii) Equivalent resistance of the network :
There are two parallel paths between points L and M. Imagine that battery of 48 Vis removed but not its internal resistance. Then, resistance of the circuit as looked into from points Land M is (Fig. 69)
R1 = R th = 24 x (6 x2) / 24 + (6 + 2) = 6 Q
(iii) Current in RL,I:
Refer Fig. 70.
I = E th / R th + RL = 36 / 6 + 30 = 1 A
Example 19. Find the current through 50 ohms resistance in the circuit shown in Fig. 71. Use Theuenin’s theorem.
Solution. To solve the problem of the network shown in Fig. 71 by Thevenin’s theorem, let R4 be assumed as disconnected as shown in Fig. 72.
With the resistance R4 disconnected, the current in the closed circuit consisting of Rh R3 and
I = E / R1 + R3 + R = 12 / 7.2 + 40 +0.8 = 0.25 A
Voltage across terminals LM = V oc = Eth = 0.25 x 40 = 10 V
The equivalent internal resistance of the network between the terminals L and M with R4
R i = R th = R2 =R3 (R1 + R / R1 + r)
= 10 + 40 (7.2 + 0.8) / 40 + 7.2 + 0.8 = 10 + 40 x 8 / 48 = 16.66 Q
Current through 50 Q (RL) resistance (Refer Fig. 73),
I = E th / R th + RL = 10 / 16.66 + 50 = 0.15 A
Example 38. The four arms of a Wheatstone bridge have the following resistances :
AB = 50 n, BC = 5 n, CD = 2 n, DA = 25 n. A galvanometer of 10 n resistance is connected across BD. If a potential difference of 20 V is maintained across AC, using Thevenin’s theorem, calculate the current through the galvanometer.
Solution. Refer Fig. 87 (i).
• After removing galvanometer from Fig. 87 (i), we get the circuit of Fig. 87 (ii).
• Let us now find the open-circuit voltage V oc (also called Thevenin voltage Eth) between Points B and D. Since ABC (as well as ADC) is a potential divider on which a voltage drop of 20 V takes place, we have
Potential of B w.r.t. C = 20 x 5 / 50 + 5 = 1.818 V
Potential of D w.r.t. C = 20 = 2 / 25 x 2 = 1.481 V
Potential difference between Band Dis V oc or Eth = 1.818 – 1.481 = 0.337 V.
- Now, remove the 20 V battery retaining its internal resistance which in this case, happens to be zero. Hence, it amounts to short-circuiting points A and C as shown in Fig. 87 (iv ).
- Next, let us find the resistance of the whole network as viewed from points Band D. It may be seen that electrically speaking points A and C have become one as shown in
Fig. 88 (i) Further it may be observed that BA is in parallel with BC and AD is in parallel with CD. Hence
RBD = 5 || 50 + 25 ||2 = 5 x 50 / 5 + 50 + 25 x 2 / 25 + 2 = 6.4 Q
• Now, so far points Band Dare concerned, the network has a voltage source of0.337 V and internal resistance R; = 6.4 Q. This Thevenin’s source is shown in Fig. 88 (iii).
• Finally, let us connect the galvanometer, which was initially removed, to this Thevenin source and calculate the current flowing through it. As seen from Fig. 88 (iv), we have 0.337 current through the galvanometer, I = 0.337 / 6.4 + 10 = 0.02 A