The circuit consists of a series of number of pairs of parallel and series
arranged resistors.
Correct responses:
- [tex]I_{BE}[/tex] = 0.5 A
Methods used for calculations to obtain the above responses
The total resistance can be calculated as follows;
Resistance between point C to D and from D to E are in series, therefore;
We have;
[tex]R_{CD}[/tex] = 5 Ω and [tex]R_{DE}[/tex] = 9 Ω are parallel
Therefore;
[tex]R_{TCE} = \mathbf{ \dfrac{1}{\dfrac{1}{R_{CE}} + \dfrac{1}{R_{CD} +R_{DE}} }}[/tex]
[tex]R_{TCE} = \dfrac{1}{\dfrac{1}{14} + \dfrac{1}{5+9} } = 7[/tex]
[tex]R_{TCE}[/tex] = 7 Ω
[tex]R_{BC}[/tex] = 11 Ω is in series with [tex]R_{CE}[/tex] both of which are parallel to [tex]R_{BE}[/tex] = 18 Ω
[tex]R_{TBE} = \mathbf{\dfrac{1}{\dfrac{1}{R_{BE}} + \dfrac{1}{R_{BC} + R_{TCE}} }}[/tex]
Therefore;
[tex]R_{TBE} = \mathbf{\dfrac{1}{\dfrac{1}{18} + \dfrac{1}{11 + 7} }} = 9[/tex]
[tex]R_{TBE}[/tex] = 9 Ω
[tex]R_{TAE} = \mathbf{\dfrac{1}{\dfrac{1}{R_{AE}} + \dfrac{1}{R_{AB} + R_{TBE}} }}[/tex]
Therefore;
[tex]R_{TAE} = \dfrac{1}{\dfrac{1}{22} + \dfrac{1}{13 + 9} } = 11[/tex]
[tex]R_{TAE}[/tex] = 11 Ω
[tex]R_T = \mathbf{R_{TAE} + R_{A_F}}[/tex]
Therefore;
- [tex]R_T[/tex] = 11 Ω + 1 Ω = 12 Ω
The current in the circuit, I, is therefore;
[tex]\displaystyle I = \frac{24 \ v}{12 \ \Omega} = \mathbf{2 \ A}[/tex]
By current divider rule, due to the equality of the parallel resistances we have;
Current through AB, [tex]I_{AB}[/tex] = [tex]\frac{1}{2} \times 2 \ A[/tex] = 1 A
Similarly;
- Current through BE, [tex]\mathbf{I_{BE}}[/tex] = [tex]\frac{1}{2}[/tex] × 1 A = 0.5 A
- Current through [tex]\mathbf{I_{A_F}}[/tex] = The circuit current, I = 2 A
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