Question

Solve question 3 Only, Be very clear/ about evry single step

Mark all currents/ voltages

assign label and polarity

redraw simplified circuits

then write KVL and KCl equations dont use values, write plain equations first then apply Ohms law.

Substitute values last. in equations

solve equations for unknowns show step by step solution for each unknown.

write clearly. evry step.

Do In Every Circuit app

Do only question 3 and attach screenshot of the outputs.

Make sure answers found match the evry circuit values.

Dont forget to attach Every Circuit Screenshot.

Its extra credit.

Please make it look neat and organised.

Do only question 3

follow all steps

This is a take-home exam. You may discuss the questions with others but the work you turn in must be yours alone. If see any similarities in two or more assignments all parties will be penalized or even worse will get zero credit no questions asked! So please try to do an honest work. 1. Mark all the currents and voltages 2. Assign proper label and polarity 3. Redraw the circuits if they can be simplified 4. Write KVL, KCL equations and apply Ohm's Law where appropriate without using actual values 5. Finally, substitute the given values in the equations Try your best to solve these equations so that you can find all the unknowns Most importantly, write legibly in an organized manner so that I can read what you have written 6. 7. 8. Clearly mark your final answers Your credit depends on the number of things you pay attention to in the list of items above. work to support them. You should take the time to test your answers with Also no credit for just answers without adequate EveryCircuit as a means of validation - if they do not match then most likely your work is flawed.

Answer 1

Hi,

Hope you are doing well.

So I'm doing only PART (3) as you requested.

(3)

So, First of all I bring your attention to the circuit given. You can notice that the current to the middle branch of the circuit is given as
So, First of all I bring your attention to the circuit given. You can notice that the current to the middle branch of the circuit is given as I_1 which will be wrong and probably, it will be I_2 as shown below. The effective resistance of the circuit is given by, 1/ R_T = 1/ R_1 r_2/ R_1 + R_2 + 1/ R_3 + R_4 therefore 1/ R_T = R_1 + R_2/ r_1. R_2 + 1/ R_3 + R_4 Given that, R_1 = 10 ohm R_2 = 15 ohm R_3 10 ohm R_4 2 Ohm therefore The effective resistance of the circuit is given by, therefore 1/ R_T = R_1 + R_2/R_1 + R_2 + 1/R_3 + r_4 = 10 ohm + 15 ohm/10 ohm middot 15 ohm + 1/ 10 ohm + 2 ohm = 25/ 150 ohm + 1/ 12 ohm 1/ R_t = 25 times 12 ohm + 150 ohm/ 150 ohm middot 12 ohm
which will be wrong and probably, it will be $I_{2}$ as shown below.

The effective resistance of the circuit is given by,

$\frac{1}{R_{T}}=\frac{1}{\frac{R_{1}.R_{2}}{R_{1}+R_{2}}}+\frac{1}{R_{3}+R_{4}}$

$\therefore \frac{1}{R_{T}}=\frac{R_{1}+R_{2}}{R_{1}.R_{2}}+\frac{1}{R_{3}+R_{4}}$

Given that,

$R_{1}=10\;\Omega$

$R_{2}=15\;\Omega$

$R_{3}=10\;\Omega$

$R_{4}=2\;\Omega$

$\therefore$ The effective resistance of the circuit is given by,

$\therefore \frac{1}{R_{T}}=\frac{R_{1}+R_{2}}{R_{1}.R_{2}}+\frac{1}{R_{3}+R_{4}}$

$=\frac{10\;\Omega+15\;\Omega}{10\;\Omega.15\;\Omega}+\frac{1}{10\;\Omega+2\;\Omega}$

$=\frac{25}{150\;\Omega}+\frac{1}{12\;\Omega}$

$\frac{1}{R_{T}}=\frac{25\times 12\;\Omega+150\;\Omega}{150\;\Omega.12\;\Omega}$

$\mathbf{\therefore R_{T}=4\;\Omega}$

We know that, According to Ohm's Law, $I_{1}$ is given by,

$I_{1}=\frac{E}{R_{T}}$

Where,

is the e.m.f of the circuit (V).

$R_{T}$ is the total/effective resistance of the circuit ( $\Omega$ ).

Given that,

e.m.f of the circuit, $E=36\;V$

Total/effective resistance of the circuit, $R_{T}=4\;\Omega$

$\therefore$ $I_{1}$ is given by,

$I_{1}=\frac{E}{R_{T}}=\frac{36\;V}{4\;\Omega}$

$\mathbf{I_{1}=9\;A}$

$\therefore$ Also, $I_{2}$ is given by,

$I_{2}=\frac{E}{\frac{R_{1}.R_{2}}{R_{1}+R_{2}}}=\frac{36\;V}{\frac{10\;\Omega\times15\;\Omega}{10\;\Omega+15\;\Omega}}$

$I_{2}=\frac{36\;V}{6\;\Omega}$

$\therefore \mathbf{I_{2}=6\;A}$

Also, $I_{3}$ is given by,

$I_{3}=\frac{E}{R_{3}+R_{4}}=\frac{36\;V}{10\;\Omega+2\;\Omega}$

$I_{3}=\frac{36\;V}{12\;\Omega}$

$\therefore \mathbf{I_{3}=3\;A}$

According to Kirchhoff's Current rule(KCL), We know that, The current reaching any junction is equal to the sum of current leaving the junction.

$\mathbf{ie;}\; I_{1}=I_{2}+I_{3}$

We have,

$\mathbf{I_{2}=6\;A}$

$\mathbf{I_{3}=3\;A}$

$\therefore I_{2}+I_{3}=6\;A+3\;A=9\;A= I_{1}$

Hence Kirchhoff's rule is proved.

From the figure, $V_{a}$ is the potential difference across the resistor $R_{4}$ , Which is given by Ohm's law as,

$V_{a}=I_{3}\times R_{4}$

Given that,

$I_{3}=3\;A$

$R_{4}=2\;\Omega$

$\therefore$ $V_{a}$ is the potential difference across the resistor $R_{4}$ , Which is given by Ohm's law as,

$V_{a}=I_{3}\times R_{4}=3\;A\times 2\; \Omega$

$\mathbf{\therefore V_{a}=6\;V}$

Hope this helped for your studies. Keep learning. Have a good day.

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