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genyjoannerubiera
25.03.2020 •
Physics
Two resistors, R 1 = 2.01 Ω R1=2.01 Ω and R 2 = 5.29 Ω R2=5.29 Ω , are connected in series to a battery with an EMF of 24.0 24.0 V and negligible internal resistance. Find the current I 1 I1 through R 1 R1 and the potential difference V 2 V2 across R 2 R2 .
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Ответ:
I1 = 3.288 A,
V2 = 17.39 V
Explanation:
The combined resistance of the two resistor connected in series is given as
R' = R1+R2 Equation 1
Where R' = Combined resistance.
Given: R1 = 2.01 Ω, R2 = 5.29 Ω
Substitute into equation 1
R' = 2.01+5.29
R' = 7.3 Ω
Using
E = I(R'+r) Equation 2
Where E = emf of the battery, I = current through the circuit, r = internal resistance.
Given: E = 24 V, R = 7.3 Ω, r = 0 Ω( Negligible)
Substitute into equation 2
24 = I(7.3)
I = 24/7.3
I = 3.288 A.
Since the resistors are connected in series, the same amount of current flows through them
Therefore,
I = I1 = 3.288 A.
Using ohm's law,
V2 = IR2 Equation 3
Where V2 = potential difference across R2 resistor.
Given: I = 3.288 A, R2 = 5.29 Ω
Substitute into equation 3
V2 = 3.288(5.29)
V2 = 17.39 V
Ответ:
chlorine's 2nd orbit also have 8 electrons
Explanation:
hope it helps you!
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