QETA004 Electrical Principles
4.1 Kirchhoff’s Laws
Name ....................
1) State Kirchhoff’s law concerning the current entering and leaving a node.
The algebraic sum of the currents entering a node is equal to Zero. (What goes in
must come out)
2) State Kirchhoff’s law concerning voltage around a closed loop
The directed sum of the electrical potential differences i.e. Voltage around any closed
circuit, is zero.
3) Given that in the circuit below, i1 is 5A and i3 is 2A what is the current i2 and does the
arrow point the correct way?
i2 is 5-2 = 3A and it flows into the node so the direction is correct.
4) In the circuit below iab is 10A icb is 2A
Draw arrows to represent the currents and determine ieb
ieb = -12A
5) In the circuit below, given that current flows in the directions shown, using
Kirchhoff’s laws concerning current entering and leaving a node, express in terms of
currents and resistances, Vbc and Vdc
Vbc is given by calculating the current out of node b which is iab-ibd
And multiplying the resistance of T by the expression for the current
So Vbc = 20(10iab- 40ibc)
6) Solve these equations for ibd
2 = 30i + 20i
ab
Eqn 1
bd
0 = 10i + 40i - 36i
ab
bd
0 = 38i +18i
bd
- 20i
ad
Eqn 2
ad
Eqn 3
ab
Here it would be useful if we could get ibd in terms of iab so that we could substitute
for iab in the first equation and hence obtain a value for iab.
To do this we need to remove iad from equation 3
To remove iad from equation 3 we could multiply equation 3 by 2 and add this
equation to equation 2.
0 = 38i +18i
bd
0 = 76i +36i
bd
- 20i
ad
- 40i
ad
ab
*2
ab
Add
0 = 10i + 40i - 36i
ab
bd
ad
0 = 116i - 30iab
bd
30iab = 116i
bd
Substitute for iab in eqn 1
2 = 116i + 20i
bd
bd
i = 2/136 = 14.7ma
bd
Eqn 1