Please help with P2-25 resistances by a thevenin equivalent. Use the voltage-div

Question

Please help with P2-25

resistances by a thevenin equivalent. Use the voltage-divider principle to predict current I. In Fig. P2.25, determine the current in the ?2-ohm resistance without using simultaneous equations. (Hint: First replace all except the 12-ohm resistance and the 42-V battery by a Thevenin equivalent.) Figure P2.25 In Fig. P2.26, with switch S closed, the ammeter A M reads 60 mA. Predict the ammeter reading with switch S open. (Hint: Does the A M reading with S closed give you a clue to the Norton equivalent?) Answer : 40 mA. 600 ohm l00hm

Explanation / Answer

sol)

open the 12 ohm resistor

find the open ckt voltage

voltage across 6 ohm resistor is

by using voltage division rule

6*35/9 = 23.33V

by using kvl

45-Voc-23.33=0

Voc=21.66V

Rth= 6 || 3= 2 ohm

replcing with thevinin equilent

Voc = Vth in series with Rth

now current in 12 ohm resistor will be

we connect 12 ohm as a load

therefore

I in 12 ohm is Vth/(Rth+12)= 21.66/14=1.54A

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