P 3.6-6 Determine the voltages v_a and v_c and the currents i_b and i_d for the circuit shown in

Figure P 3.6-6.

Answer: v_a = –2 V, v_c = 6 V, i_b = –16 mA, and i_d = 2 mA

Figure P 3.6-6

We can get i_d from current division.

Now we can calculate the voltage drop over the 10 kΩ resistor which must be the same as the 2500 Ω since they are in parallel.

So since the 10 mA supply and the 10 kΩ are in parallel with the 18 V supply, we must have

For the 2 kΩ and 1 kΩ resistor in series, we know they have 18 V across them so the current through them is

Therefore v_c is found from

Finally i_b can be found from a junction rule

- sign due to current goes opposite to how i_b was drawn.

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This page last updated on January 20, 2019