Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 2-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified.

(a) Two resistors are connected in series across a 2-V battery. Let R1 = 10 ? and R2 = 9 ?.
Req = ?
I = A
?V1 = V
?V2 = V

(b) Add a third resistor to the circuit in series. Let R1 = 10 ?, R2 = 9 ?, and R3 = 3 ?.
Req = ?
I = A
?V1 = V
?V2 = V
?V3 = V

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lublana lublana · Answer 1 · 2019-11-06T11:44:16+01:00

Answer with Explanation:

We are given that

Potential difference =2V

a.[tex]R_1=10 ohm,R_2=9 ohm[/tex]

We know that in series

[tex]R_{eq}=R_1+R_2[/tex]

[tex]R_{eq}=10+9=19 ohm[/tex]

[tex]I=\frac{V}{R}[/tex]

[tex]I=\frac{2}{19}=0.105 A[/tex]

[tex]V_1=IR_1[/tex]

[tex]V_1=\frac{2}{19}\times 10=1.05 V[/tex]

[tex]V_2=\frac{2}{19}\times 9=0.95 V[/tex]

b.[tex]R_1=10 ohm , R_2=9 ohm ,R_3=3 ohm[/tex]

[tex]R_{eq}=10+9+3=22\Omega[/tex]

[tex]R_{eq}=22\Omega[/tex]

[tex]I=\frac{2}{22}=\frac{1}{11}=0.09 A[/tex]

[tex]V_1=10(0.09)=0.9 V[/tex]

[tex]V_2=9(0.09)=0.81 V[/tex]

[tex]V_3=3(0.09)=0.27 V[/tex]