Chapter 17 & 18

Direct Current & Circuits

Example 18-7 Solution

First since there is a capacitor in the circuit we need to look at that part first. When it says steady-state there is no current through that part of the circuit so we can ignore it.

We have two nodes again at c and f. Both nodes produce identical equations so we will use the one at point c:

(1)

Lets use loops defcd and cfgbc for the other two equations:

(2) defcd

(3) cfgbc

If we solve equation (1) for we get and then substitute this into equation (3) and simplify:

(4)

Then subtracting equation (4) from equation (2) will eliminate , then solve for :

Using equation (3) and equation (1) gives:

= 1.38 A and = 1.02 A

(b) We can apply the loop rule for the capacitor loop to find the potential difference across the capacitor.

because the charge on the capacitor is:

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