A student in the laboratory connects a 17 Ω resistor, a 22 Ω resistor, and a 38 Ω resistor in series and then?
A student in the laboratory connects a 17 Ω resistor, a 22 Ω resistor, and a 38 Ω resistor in series and then?
A student in the laboratory connects a 17 Ω resistor, a 22 Ω resistor, and a 38 Ω resistor in series and then connects the arrangement to a 35-V dc source.
(a) How much current is in the circuit? __________A
(b) How much power is expended in the circuit?__________W
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Pretty straightforward question so:
a. Total Resistance / Equivalent Resistance = (17 + 22 + 38)Ω = 77 Ω
Applied Voltage = 35 V
Ohm's law states V = IR
Hence, I = V / R
I = 35 / 77 = 0.45... A
b. Power can be calculated as follows:
P = I^2R
P = IV
P = V^2 / R
Anyone can work here so let us use the simplest one which is P = IV
P = (0.45)(35)
P = 15.75 W
*Note that when u use some of the equations some numbers after the decimal point will differ because we can't get an accurate current but the whole number will be the same*
in sequence, R equivalence is purely the sum of the resistors, subsequently: 12+30+33=seventy 5 ohms. a. modern-day = Voltage/ Resistance = 40V/ seventy 5 ohms = 0.fifty 3 amperes. b. ability = V. I = 40V x 0.fifty 3 A = 21.3 watts