P28 070

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70. In the steady state situation, the capacitor voltage will equal the voltage across the 15 kΩ resistor:

V

0

= (15 kΩ)



20 V

10 kΩ + 15 kΩ



= 12 V .

Now, multiplying Eq. 28-36 by the capacitance leads to V = V

0

e

−t/RC

describing the voltage across

the capacitor (and across the R = 15 kΩ resistor)after the switch is opened (at t = 0). Thus, with
t = 0.00400 s, we obtain

V = (12)e

0.004/(15000)(0.4×10

6

)

= 6.16 V .

Therefore, using Ohm’s law, the current through the 15 kΩ resistor is 6.16/15000 = 4.11

× 10

4

A.


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