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HannahPandaidn

a) [tex]E_1\:=\:40\:V[/tex]

b) [tex]E_G\:=\:120\:V[/tex]

c) [tex]R_T \:= \: 12 Ω\:[/tex]

d) [tex]R_2 \:= \: 8 Ω\:[/tex]

Explanation:

Given:

[tex]I_T\:=\:10 \:A[/tex]

[tex]R_1\:=\:4 \:Ω[/tex]

[tex]E_2\:=\:80 \:V[/tex]

Required:

a) [tex]E_1[/tex]

b) [tex]E_G[/tex]

c) [tex]R_T[/tex]

d) [tex]R_2[/tex]

Equation:

Since the circuit is a series connection:

1) Current is same throughout the connection

[tex]I_T \: = \:I_1 \: = \:I_2\: = \: I_3 \: = \:... = \: I_nI [/tex]

2) Total Voltage is the sum of all the voltages in the connnection

[tex]E_T \: = \: E_1 \: + \: E_2\: + \: E_3 \: + \:... \:+ \: E_n[/tex]

3) Resistance is the sum of all the voltages in the connnection

[tex]R_T \: = R_1 \: + \: R_2\: + \: R_3 \: + \: ...\: + \: R_n[/tex]

Ohm's law states that

E = IR

where: E - voltage, V

I - current, A

R - resitance, Ω

Solution:

a) [tex]E_1[/tex]

Solve for [tex] E_1[/tex]

[tex] E_1 \:= \:I_1 R_1[/tex]

[tex] E_1 \:= \:(10\:A)(4 \:Ω)[/tex]

[tex] E_1\:= \:40 V[/tex]

b) [tex]E_G[/tex]

[tex] E_G = E_1\:+\: E_2[/tex]

[tex] E_G = 40\:V\:+\:80\:V[/tex]

[tex] E_G = 120\:V[/tex]

c) [tex]R_T[/tex]

Since current is same throughout the connection,

[tex]I_T \: = \:I_1 \: = \:I_2\: = \:10 \: A[/tex]

[tex] E_G \:=\: I_T R_T[/tex]

Substitute [tex] E_T and I_T[/tex]

[tex] 120\:V \:= \: (10\:A) R_T[/tex]

[tex] 120\:V \:= \: (10\:A) R_T[/tex]

[tex] R_T \:= \: \frac{120\:V}{10\:A}[/tex]

[tex] R_T \:= \: 12\: Ω[/tex]

d) [tex]R_2[/tex]

[tex] R_T \:= \:R_1 \:+\: R_2 [/tex]

[tex] 12\:Ω \:= \: 4 Ω\:+\:R_2[/tex]

[tex] R_2 \:= \: 12 Ω\:- \:4\:Ω[/tex]

[tex] R_2 \:= \: 8 Ω\:[/tex]

Final Answer:

a) [tex]E_1\:=\:40\:V[/tex]

b) [tex]E_G\:=\:120\:V[/tex]

c) [tex]R_T \:= \: 12 Ω\:[/tex]

d) [tex]R_2 \:= \: 8 Ω\:[/tex]

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