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Find the emf E1 in the circuit of the figure (Figure 1) .
Find the emf E2 in the circuit of the figure.
Find the potential difference of point b relative to point a
This problem can be solved using the concepts of Ohm’s law and Kirchhoff’s voltage laws, Electromotive force, Ohms law and Kirchhoff’s power law.
Refer to the circuit diagram in the question. Find the [katex]{\varepsilon _1}[/katex] values later.
The following is the potential difference between point and point , a:
[katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]
Here, [katex]{V_{\rm{b}}}[/katex]
The Kirchhoff’s voltage law states that the sum of all voltages at point A is zero.
To write the voltage inside the loop, you need to multiply the current and resistor by the sum of their resistance.
[katex]V = IR[/katex]
I here is the current, and R the resistance.
(1)
Below is an illustration of Kirchhoff’s voltage loop 1 & 2:
The above illustration shows that there are loops. Loop 1 is the top loop, while loop 2 is the larger loop.
The KVL loop 1 law states that the following expressions are allowed:
[katex]20.{\rm{0 V}} – {\rm{1}}{\rm{.00 A}}\left( {1.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.00{\rm{ }}\Omega } \right) – {\varepsilon _1} – 6.00{\rm{ A}}\left( {1.00{\rm{ }}\Omega } \right) = 0[/katex]
Solve for [katex]{\varepsilon _1}[/katex]
[katex]\begin{array}{c}\\20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {4.0{\rm{ }}\Omega } \right) + 1.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – {\varepsilon _1} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} – 1.00{\rm{ V}} + {\rm{4}}{\rm{.00 V}} + 1.{\rm{00 V}} – {\varepsilon _1} – 6.00{\rm{ V = 0}}\\\\{\varepsilon _1} = 18{\rm{ V}}\\\end{array}[/katex]
[(1)]
(1)
[(1)]
(2)
The KVL loop law states that the following expressions are allowed:
[katex]20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) – {\varepsilon _2} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0[/katex]
Solve for [katex]{\varepsilon _2}[/katex]
[katex]\begin{array}{c}\\20{\rm{ V}} – {\rm{1}}{\rm{.00 A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) – 2.00{\rm{ A}}\left( {2.0{\rm{ }}\Omega } \right) – {\varepsilon _2} – 6.00{\rm{ A}}\left( {1.0{\rm{ }}\Omega } \right) = 0\\\\20.{\rm{0 V}} – 1.00{\rm{ V}} – 2.{\rm{00 V}} – 4.{\rm{00 V}} – {\varepsilon _2} – 6.00{\rm{ V = 0}}\\\\{\varepsilon _2} = 7.0{\rm{ V}}\\\end{array}[/katex]
[(2)]
(2)
[(2)]
(3)
Use the KVL laws to complete the loop and then write the following expression:
[katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]0
Solve for [katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]1
[katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]2
Substitute 18.0 V to [katex]{\varepsilon _1}[/katex]
[katex]{V_{{\rm{ab}}}} = {V_{\rm{b}}} – {V_{\rm{a}}}[/katex]4
[(3)]
(3)
[(3)]
Ans:
The [katex]{\varepsilon _1}[/katex] magnitude
The [katex]{\varepsilon _2}[/katex] magnitude
The potential difference between point b and point a is equal or -13.0 V.