Consider a circuit with two resistors in parallel R_1 = 10 ohm and R_2 = 5 ohm.A) Determine the total resistance of the circuit.B) Calculate the total current flowing through the battery.C) What are the current and voltage through each resistor?D) Repeat the calculations for a series configuration.

Respuesta :

Answer:

Explanation:

Given

[tex]R_1=10 \Omega [/tex]

[tex]R_2=5 \Omega [/tex]

when resistance in Parallel

[tex]\frac{1}{R_{p}}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]

[tex]R_p=\frac{R_1R_2}{R_1+R_2}[/tex]

[tex]R_p=\frac{10}{3}[/tex]

Suppose V is voltage of battery

Total Current [tex]i=\frac{3V}{10}[/tex]

Since Circuit is Parallel therefore Voltage across both resistor is same

[tex]V=i_1R_1=i_2R_2[/tex]

and [tex]i_1+i_2=i[/tex]

[tex]i_1+i_1\cdot \frac{R_1}{R_2}=i[/tex]

[tex]i_1(1+\frac{10}{5})=\frac{3V}{10}[/tex]

[tex]i_1=\frac{V}{10}[/tex]

[tex]i_2=\frac{2V}{10}[/tex]

(b) When Circuit is in series

[tex]R_s=R_1+R_2[/tex]

[tex]R_s=10+5=15 \Omega [/tex]

since circuit is in Series therefore current is same in both resistor

Current [tex]i=\frac{V}{15} A[/tex]

Voltage drop across [tex]R_1=i\times R_1[/tex]

[tex]V_1=\frac{V}{15}\times 10=\frac{2V}{3}[/tex]

[tex]V_2=\frac{V}{15}\times 5=\frac{V}{3}[/tex]              

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