[tex] E_1 \: = 32 \: V[/tex]
[tex] E_2 \: = 96 \: V[/tex]
Explanation:
Given:
[tex] R_1 \: = \: 10\: Ω [/tex]
[tex] R_2 \: = \: 30\: Ω [/tex]
[tex] E_T \: = \: 128\: V [/tex]
Required:
[tex] E_1 [/tex]
[tex] E_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_n[/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) Resistancet 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:
First step is to solve for Total Resistance
[tex] R_T \: = R_1 \: + \: R_2[/tex]
[tex] R_T \: = \: 10\: Ω \: + \: 30\: Ω[/tex]
[tex] R_T \: = \: 40\: Ω[/tex]
Solve for the Total Current
[tex] E_T \: = (I_T)(R_T)[/tex]
[tex] I_T \: = \frac{E_T}{R_T} [/tex]
[tex] I_T \: = \frac{E_T}{R_T} [/tex]
[tex] I_T \: = \frac{128\: V}{40\: Ω} [/tex]
[tex] I_T \: = 3.2\: A [/tex]
Solve for [tex] I_1 and I_2 [/tex]
Since [tex] I_T \: =\: I_1 \: = \: I_2\: =\: I_3 \: =\:... = \:I_n[/tex]
[tex] I_T \: =\: I_1 \: = \: I_2\: =\:3.2 \: A [/tex]
Solve for [tex] E_1 [/tex]
[tex] E_1 \: = (I_1)(R_1)[/tex]
[tex] E_1 \: = (3.2 \: A)( 10\: Ω)[/tex]
[tex] E_1 \: = 32 \: V[/tex]
Solve for [tex] E_2 [/tex]
[tex] E_2 \: = (I_2)(R_2)[/tex]
[tex] E_2 \: = (3.2 \: A)( 30\: Ω)[/tex]
[tex] E_2 \: = 96 \: V[/tex]
Final Answer
[tex] E_1 \: = 32 \: V[/tex]
[tex] E_2 \: = 96 \: V[/tex]
