Answer :
Answer:
I = 3.3 A
V2 = 17.4 V
Explanation:
a)
- In a circuit series, the current is the same through any point of the circuit.
- Assuming the resistors are in the linear zone of operation, we can apply Ohm's Law to both resistors, as follows:
[tex]V = I* r_{eq} = V_{R1} + V_{R2} = (I*R_{1}) + (I*R_{2}) = I * (R_{1} + R_{2})\\ \\ R_{eq} = R_{1} + R_{2}[/tex]
- Therefore, we can find the current I as follows:
[tex]I =\frac{V}{R_{eq} } = \frac{24.0 V}{7.3 \Omega} = 3.3 A[/tex]
b)
- Applying Ohm's law to R2, we can find the voltage through R2 as follows:
[tex]V_{R2} = I* R_{2} = 3.3 A * 5.29 \Omega = 17.4 V[/tex]
Answer:
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