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]
