At the end of this lesson, the students should be able to:
• State Kirchhoff’’s laws
• Explain Kirchhoff’s laws
• Solve simple problems on Kirchhoff’s laws
2.10 KIRCHHOFF’S LAWS
Kirchhoff’s laws were developed in 1847 by the German physicist G.R. Kirchhoff. These laws are two in numbers and they are formally known as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL).
Kirchhoff’s current law Kirchhoff’s current law state that the algebraic sum of currents entering a node is zero.
By this law for purposes of explanation, we may regard currents entering a node as positive, where currents leaving the node as negative. Considering the node in Fig 8.1 and applying KCL, we get
i1 + (-i2) + (-i3) + i4 + i5 = 0
(8.1)
or
i1 + i4 + i5 = i2 + i3
(8.2)
From equation (8.2) we notice that the sum of current entering a node is equal to the sum of the current leaving the node. This is an alternative way of stating KCL.
Example 8.1
Find the value of I in Fig 8.2
Solution
In order to find the value of I, the following equation can be formed.
Currents towards P = currents flowing from P. ∴ 6 + 4 + 5 = I + 8, ⇒ I = 7A
Kirchhoff’s voltage law
Kirchhoff’s voltage law state that in a closed circuit, the algebraic sum of the e.m.f.s is equal to the algebraic sum of the voltage drops.
Mathematically this can be stated as:
± E1 + ±E2 = ± I1R1 ± I2R2 ± I3R3,
if there are two sources of e.m.f and three resistors R1, R2 and R3 In general, ∑ E = ∑ IR
Example 8.2
Determine the value of Vab in the circuit shown in Fig 8.3 (a) and (b) using KVL.
Solution
For Fig 8.3(a), applying KVL, we get
-2V + 3V – 6V – Vab = 0
∴ Vab = -5V
For Fig 8.3 (b) applying KVL, we get
-2V + 3V + 6V – Vab = 0
∴ Vab = 7V
Example 8.3
Applying KCL and KVL to the circuit of Fig. 8.4, write down the equation pertaining to loop I and II respectively,
N.B. we are working under the assumption of fixed direction of I1, I2 and I3 as shown in Fig 8.4
Applying KVL to loop I, we get
-E1 = R1I1 + R2I2 Applying KVL to loop II, we get E2 = -R2I2 + R3I3
From KCL we known that at node A
I1 = I2 + I3 or I3 = I1 – I2
Putting this value of I3 into the equation involving E2, we get
E2 = -R2I2 + R3 (I1 – I2)
E2 = R2I2 – (R2 + R3) I2
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