IP Three resistors, R1 = 15 Ω , R2 = 62 Ω , and R3=R, are...
Homework Answers
Solution- From the question we have
R1 = 15 Ω :R2 = 62 Ω :R3 =R
voltage V = 24 V
current I =0.15A
The three capacitors are
connected in series combination
I = 24/15+62+R
=>0.15 = 24/15+62+R
=>R = 24- 0.15(15+62)/(0.15)
=83 Ω
b) V(15)= 0.15*15= 2.25 V
V(62)= 0.15*62=9.3 V
V(83)= 0.15*83= 12.45 V
.
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connected in parallel, and this parallel combination is then
connected in series with a 100 Ω resistor, R3 as shown
below
If the total resistance of the circuit is 300 Ω what must be the
resistance of R1 and R2? If the circuit is
connected to a 30V battery, what would the current through
R1? What voltage is a dropped across R3?
100 S
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circuit as shown below. Calculate the voltage across R2.
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Volts)
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