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By Jerald G. Graeme

This easy-to-use publication should be a pragmatic reference for circuit designers and clients of operational amplifiers in lots of various engineering and clinical fields

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Extra resources for Designing with operational amplifiers: applications alternatives

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1-17. CHAPTER 2 Kirchhoff’s Laws and Resistance In this chapter we will encounter two laws that are general enough to apply to any circuit. The first of these, Kirchhoff’s current law, is a result of the conservation of charge and tells us that the sum of currents at a connection point in a circuit must vanish. The second law, which derives from the conservation of energy, is Kirchhoff’s voltage law. This law tells us that the sum of voltages in a closed path in a circuit must vanish. After describing these laws in more detail, we will consider the concept of resistance and meet our first real element in circuit analysis, the resistor.

KCL at this node is then written as i = i1 − i2 = 0 Let’s apply KCL to a more substantial example. EXAMPLE 2-1 Consider the circuit shown in Fig. 2-2. If i 1 = 3 A, i 3 = 5 A, i 4 = 6 A, and i 5 = 1 A, find i 2 . SOLUTION KCL tells us that the sum of the currents at the node shown in Fig. 2-2 must vanish. That is, in = 0 Taking + for currents entering the node and − for currents leaving the node, KCL gives us i1 − i2 − i3 + i4 − i5 = 0 Solving for i 2 , i2 = i1 − i3 + i4 − i5 = 3 − 5 + 6 − 1 = 3 A EXAMPLE 2-2 Consider the node shown in Fig.

1-13. We repeat the calculations we did for the circuit element shown in Fig. 1-12. This time, looking at Fig. 1-13, we need to reverse the sign of the current. If v(t) = 5 V and i(t) = 3 A, the power for the element in Fig. 1-13 is p = (5 V)(−3 A) = −15 W Since the power is negative, the element delivers power. On the other hand, suppose that i(t) = −3 A. Then p = (5 V) (− (−3A)) = +15 W In other words, the circuit absorbs power. EXAMPLE 1-8 Determine the power supplied or absorbed for each element in the circuit shown in Fig.

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