I need help in solving these questions. I've tried to use Kirchhoff's law but this hasn't worked.
I have been having a lot of difficulty with the resistance and current calculations, would someone be able to help show me where I have made mistakes.
Electrical – Calculating resistance, voltage and current in parallel circuit
homeworkparallel
Related Solutions
Well, I'm studying electrical engineering right now and I can tell you that such jumps as you described take around two years of lectures at my university.
First thing which is important is to know which elements are passive and which are active. Then you need to know which elements are linear and which aren't. Next step is to get equivalent schematics for elements which you have and to see how they behave.
For example, let's take the switch. In off state, it functions as an open circuit, while in on state it functions as short circuit. Next, if you have sensitive equipment, you'll be able to notice that the switch isn't actually short circuit because it has some resistance, but that it's very low. Now let's take a look at the diode. Diode isn't linear component, so it doesn't have resistance in the classical sense in which for example resistors have. Instead there's the V-I curve of the diode. On a resistor, it's a linear function and we can use resistance as its characteristic, but on diode, it looks exponential.
As you can see from the image, certain voltage is needed for diode to start working properly and when you trigger the switch, that voltage disappears. That means, that the "resistance" of the diode just became huge. To get a feeling for this, use the parallel resistor calculation for say 1 mΩ resistor and 1MΩ resistor and take a look how much current goes through each of them. This is the way the circuit you mentioned behaves.
I left out the units.
total resistance \$R\$
\$R_1\$ (between a and c) is in parallel to the rest of the resistors. The rest being \$R_2\$ parallel to \$R_3\$ (between a and b) in row to \$R_4\$ in parallel to \$R_5\$ (between b and c)
The general rule for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel:
$$R_a||R_b=\frac{R_aR_b}{R_a+R_b}$$
I gave it a try:
$$ \begin{align} R & = R_1||(R_2||R_3 +R_4||R_5)\\ & = R_1||(R_{23} + R_{45})\\ & = \frac{R_1(R_{23} + R_{45})}{R_1+R_{23} + R_{45}}\\ R_{23} & = \frac{R_2R_3}{R_2+R_3} = \frac{20.46}{9.5}\\ R_{45} & = \frac{R_4R_5}{R_4+R_5} = \frac{56}{15.6}\\ R & = \frac{1(\frac{20.46}{9.5} + \frac{56}{15.6})}{1+\frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5}}{\frac{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{15.6 \times 9.5}} =\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5} \\ R & \approx 0.85 \end{align} $$
voltage \$U_{ab}\$ via voltage divider
I use U for voltage, not V.
\$R_1\$ being in parallel to the rest of the resistors means that there's the same voltage over both of them. The voltage divider divides the voltage \$U_{ac}\$ into \$U_{ab} + U_{ac}\$
The general rule for a voltage divider for two resistors in \$R_{ab}\$ and \$R_{bc}\$ in row: $$\frac{U_{ab}}{U_{ac}}=\frac{R_{ab}}{R_{ac}}=\frac{R_{ab}}{R_{ab} + R_{bc}}$$
I gave it a try: $$ \begin{align} \frac{U_{ab}}{U_{ac}} &= \frac{R_{23}}{R_{23} + R_{45}} \\ &= \frac{ \frac{20.46}{9.5}}{ \frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{ \frac{20.46}{9.5}}{ \frac{20.46 \times 15.6 + 56 \times 9.5}{9.5 \times 15.6}} = \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5}\\ &\approx 0.3750 \\ U_{ac} &= 5 \\ U_{ab} &= \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 1.8750 \end{align} $$
current \$I_{R_2}\$ via current divider
The current divider divides the current \$I_{ab}\$ into \$I_{R_2} + I_{R_3}\$
The general rule for a current divider for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel: $$\frac{I_{a}}{I_{a} + I_{b}}=\frac{I_{a}}{I_{ab}}=\frac{R_{a} || R_{b}}{R_{a}}= \frac{R_aR_b}{R_a(R_a +R_b)}=\frac{R_{b}}{R_{a} + R_{b}}$$
The general rule for resistance, voltage and current (ohm's law) : $$R = \frac{U}{I} \iff I = \frac{U}{R}$$
I gave it a try:
$$ \begin{align} I_{ab} &= \frac{U_{ab}}{R_{ab}} = \frac{U_{ab}}{R_{23}}\\ &= \frac{\frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5}{\frac{20.46}{9.5}} = \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.8706\\ \frac{I_{R_2}}{I_{ab}} &= \frac{R_3}{R_2 + R_3} = \frac{3.3}{9.5} \\ &\approx 3.4747 \\ I_{R_2} &= \frac{R_3}{R_2 + R_3} \times I_{ab} = \frac{3.3}{9.5} \times \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5 = \frac{15.6 \times 3.3}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.3024\\ \end{align} $$
power \$P_{R_1}\$
Given the overall voltage \$U_{ac}\$ and the resistance \$R_1\$ the power \$P_{R_1}\$ can be calculated.
The general rule for power: $$P = U \times I$$ With ohm's law: $$P =\frac{U^2}{R}$$
I gave it a try: $$ \begin{align} P_{R_1} &=\frac{U_{ac}^2}{R} =\frac{5^2}{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}} =\frac{25 \times 15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{20.46 \times 15.6 + 56 \times 9.5} \\ &\approx 5.3528 \end{align} $$
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Best Answer
You're calculating total resistance incorrectly. Parallel resistance is \$ \frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \cdots \$
Where you get the \$(1+1+1)/3 \$ from, I'm not entirely sure.
\${R_{1}}\$ and \${R_{2}}\$ are in series. These are in parallel with \${R_{3}}\$ and \${R_{4}}\$ which are in series.
First thing you want to do is simplify each branch into a single resistor. As \${R_{1}}\$ and \${R_{2}}\$ are in series we can simply add them. So the total resistance in our first branch is \$4\Omega\$.
If we do the same with our second branch (So \${R_{3}} + {R_{4}}\$) then we can see the total resistance in our second branch is also \$4\Omega\$.
Now we have a circuit that looks like this:
(By the way, whichever teacher drew this needs to go back to school themselves imo. Nobody would ever draw a parallel circuit like this, Google 'parallel resistor circuit to see my point')
Now that you've got this simplified circuit, finding parallel resistance is a lot simpler. Using the equation \$ \frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \cdots \$ we can put our values in to get \$\frac{1}{R_{T}} = \frac{1}{4} + \frac{1}{4}\$, this comes out to be \$\frac{1}{R_{T}} = 0.5\$.
However, we don't want \$\frac{1}{R_{T}}\$ we just want \${R_{T}}\$ so we have to invert the equation like so. \$\frac{1}{R_{T}} = 0.5 \rightarrow {R_{T}}=\frac{1}{0.5}\$. This gives a value of \${R_{T}} = 2\Omega\$.
(As there is only 2 resistances you can also use \$\frac{R_{a}\times R_{b}}{R_{a}+R_{b}} \$ but it's personal choice)
You should be able to do the rest of the questions now with ease if you follow Krishn's advice in the other answer