Whenever the reference direction for the current in an element is in the
direction of the reference voltage drop across the element (as in Fig. 1.5), use
a positive sign in any expression that relates the voltage to the current.
Otherwise, use a negative sign.
We apply this sign convention in all the analyses that follow. Our purpose for
introducing it even before we have introduced the different types of basic
circuit elements is to emphasize that selecting polarity references is not a
function either of the basic elements or the type of interconnections made
with the basic elements. We apply and interpret the passive sign convention
for power calculations in future posts.
Example 1.2 illustrates one use of the equation defining current.
Example 1.2 Relating Current and
Charge
No charge exists at the upper terminal of the element in Fig. 1.5 for t<0. At
t=0, a 5 A current begins to flow into the upper terminal.
direction of the reference voltage drop across the element (as in Fig. 1.5), use
a positive sign in any expression that relates the voltage to the current.
Otherwise, use a negative sign.
We apply this sign convention in all the analyses that follow. Our purpose for
introducing it even before we have introduced the different types of basic
circuit elements is to emphasize that selecting polarity references is not a
function either of the basic elements or the type of interconnections made
with the basic elements. We apply and interpret the passive sign convention
for power calculations in future posts.
Example 1.2 illustrates one use of the equation defining current.
Example 1.2 Relating Current and
Charge
No charge exists at the upper terminal of the element in Fig. 1.5 for t<0. At
t=0, a 5 A current begins to flow into the upper terminal.
- Derive the expression for the charge accumulating at the upper terminal of the element for t>0.
- If the current is stopped after 10 seconds, how much charge has accumulated at the upper terminal?
- From the definition of current given in Eq. 1.2, the expression for charge accumulation due to current flow
isq(t)=∫0ti(x)dx.
Therefore,
q(t)=∫0t5dx=5x|0t=5t−5(0)=5t C for t>0. - The total charge that accumulates at the upper terminal in 10 seconds due to a 5 A current is q(10)=5(10)=50 C.
Assessment Problems
Objective 2—Know and be able to use the definitions of voltage and
current
1. 1.3 The current at the terminals of the element in Fig. 1.5 is
i=0,t<0;i=20e−5000t, A,t≥0;
Calculate the total charge (in microcoulombs) entering the element at its
upper terminal.
Answer: 4000 μC.
2. 1.4 The expression for the charge entering the upper terminal of Fig. 1.5
is
q=1α2−(tα+1α2)e−αt C.
Find the maximum value of the current entering the terminal if
α=0.03679 s−1.
Answer: 10 A.
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