A 20.0-Ω resistor and a 35.0-Ω resistor in series are connected to a 120-V dc source.?
A 20.0-Ω resistor and a 35.0-Ω resistor in series are connected to a 120-V dc source.
(a) What is (are) the current(s) through the resistors?
current through the 20.0-Ω resistor:_________A
current through the 35.0-Ω resistor:_________A
(b) What is the voltage drop across each resistor?
voltage drop across the 20.0-Ω resistor:_________V
voltage drop across the 35.0-Ω resistor:_________ V
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Verified answer
R(total)=R1+R2=55.0 ohm
v=iR
120=i×55
I=120/55=24/11(A)
Current through the resistors are equal
Also equal to the current through the total resistor
Since they are in series.
i1=i2=24/11(A)
V1=i1×R1=24/11×20.0=480/11 (v)
V2=i2×R2=24/11×35=840/11 (v)
Or v2=120-480/11=840/11(v)
allow R = resistance of light bulb entire resistance of light bulb circuit = one hundred forty 5 + R voltage of light bulb circuit = 120 capacity presented to mild bulb = 24.2 w = I²R {the position I = modern-day through mild bulb circuit} R = 24.2/I² IR + 145I = 120 I(24.2/I²) + 145I = 120 24.2/I +145I = 120 24.2/I = 120 - 145I 24.2 = 120I - 145I² 0 = 145I² + 120I - 24.2 sparkling up quadratic for I: I = 0.168 amp ANS-a million I = - 0.995 amp ANS-2