In the circuit below, a parallel combination of a 10.0-Ohm resistor and a 7.0 mH

Question

In the circuit below, a parallel combination of a 10.0-Ohm resistor and a 7.0 mH inductor is connected in series with an R = 8.0-Ohm resistor, a 6.2-V dc battery, and a switch. What are the voltage across the 8.0-Ohm resistor and the 10.0-Ohm resistor, respectively, immediately after the switch is closed V_8.0 Ohm = 0.41 times V V_10.0 Ohm = _____ v what are the voltages across each of these resistors after the switch has been closed for a long time? V_8.0 Ohm = ____ v V_10.0 Ohm = ____ v What is the current in the 7.0-mH inductor after the switch has been closed for a long time? ______ A

Explanation / Answer

(a)
Inductor acts as an open cicuit, immediately after switch is closed !!
So,
I = V/(8.0 +10.0) = 6.2 / (8.0 + 10.0) = 0.344 A
Volatge across 8.0 = 0.344 * 8.0 = 2.75 V
Volatge across 10.0 = 0.344 * 10.0 = 3.45 V

(b)
Inductor acts as a Short cicuit, after switch is open for a long time !!
Volatge across 8.0 = 6.2 V
Volatge across 10.0 = 0 V

(c)
As Inductor acts as a Short cicuit, after switch is open for a long time, all the current will pass throuh inductor

I = V/8.0 = 6.2/8.0
I = 0.775 A

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