From the circuit calculate Ib. It looks pretty simple but I\'m not getting it. P

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Ok. since you must have ib + 100 ib flowing through the 1Kresistor, you know 101 ib is flowing through thisresistor. Also, by applying KCL at the node between the 100K and 10Kresistors, and if we choose an arbitrary current going through the100K resistor, call it i, then the current going through the 10Kresistor must be i - ib (to satisfy KCL) Now we can apply KVL in 2 spots using these 2 unknown variables,use substitution, and solve. the equations will be: (loop from ground to ground) -> 10K * (i -ib) - 0.7V - 101 ib *1000 = 0   (1) (loop from 12V source to ground running through 1K resistor) -> 12V - 100K * i - 0.7 - 1K*101 ib =0 by solving i in terms of ib from (1), and then plugging into (2)and solving for ib, I get ib = 3.55 uA and i = 0.109 mA Go over this yourself and see if it makes sense to you. I can notgarantee that this is correct so make sure you go over it andsee. I think it's right though. hope this helps.

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