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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