Electric Circuit Diagrams (Only Kidding)

December 20, 2022July 24, 2023

202212201748 Solution to 2020-DSE-PHY-IA-23

Three identical resistors, a battery of negligible internal resistance, and an ideal voltmeter are connected to form Circuits (a) and (b) respectively.

Given that the voltmeter reading is $8\,\mathrm{V}$ in Circuit (a), what is the voltmeter reading in Circuit (b)?

Roughwork.

Redrawing a labelled diagram for (a),

and noting that

$\begin{aligned} I_1=I_2 & = I/2 \\ I_3 & = 0 \\ I_4 & = I \\ R_\textrm{V}\parallel R & = R \\ R\parallel R & = R/2 \\ R_{\textrm{eq}} & = 3R/2 \\ \end{aligned}$

we are about to write

$\begin{aligned} I_4R & =8\,\mathrm{V}\textrm{ is given} \\ \mathcal{E} & =IR_{\textrm{eq}}\\ & = I\bigg(\frac{3R}{2}\bigg) \\ & = \frac{3}{2}(8) \\ & = 12\,\mathrm{V} \\ \end{aligned}$

and obtain that the battery has an emf $\mathcal{E}$ of $12\,\mathrm{V}$ . Likewise for (b),

where

$\begin{aligned} I_2 & = I_4 \\ I_3 & = 0 \\ R_\textrm{V}\parallel R & = R \\ R+R_\textrm{V}\parallel R & = 2R \\ R\parallel (R+R_\textrm{V}\parallel R) & = 2R/3 \\ 12= \mathcal{E} & = I\bigg(\frac{2R}{3}\bigg) \\ \Longrightarrow\enspace IR & = 18 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \begin{cases} I_1R =I_2(2R) \\ I_1+I_2 = 18/R \\ \end{cases} \\ & \Longrightarrow \begin{cases} I_1 = 12/R\\ I_2 = 6/R\\ \end{cases} \end{aligned}$

being now asked for $V=I_4R$ , the voltmeter reading, it is left to the reader.

December 7, 2022December 8, 2022

202212071217 Solution to 1971-HL-PHY-I-7

The figure below shows a three-dimensional network in the form of a pyramid, in which $A$ is the apex, and $BCDE$ the square base.

Each of the eight edges of the pyramid is a wire of resistance $1\,\mathrm{\Omega}$ . A $12\,\mathrm{V}$ battery with internal resistance of $0.1\,\mathrm{\Omega}$ is connected across $B$ and $D$ . Calculate

(a) the power input of the network,
(b) the terminal voltage across the battery when current flows through the network, and
(c) the potential at the points $B$ , $C$ , $D$ , and $E$ if the apex $A$ is earthed.

Roughwork.

Draw the circuit.

Label the potential.

Straighten the main.

Calculate the equivalent.

$\begin{aligned} R_{\textrm{eq}} & = \big( R\parallel R\parallel R\big) + \big( R\parallel R\parallel R\big) \\ & = \frac{1}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}\times 2 \\ & = \frac{2R}{3} \\ & = \frac{2(1)}{3} \\ & = \frac{2}{3}\,\mathrm{\Omega} \\ \end{aligned}$

This problem is not to be attempted.

February 19, 2019January 17, 2023

201902190827 Electric circuit diagrams (Elementary) Q13

The blogger claims no originality of his idea here.

We are given the following diagram:

Circuit0013

We would like to classify the nodes by the electric potential there. We label the nodes with $a$ , $b$ , $c$ , etc. where the potential at $a$ is larger than that at $b$ , the potential at $b$ is larger than that at $c$ , etc. Therefore we have the following diagram:

Circuit0013Sol001

Aligning $a$ , $b$ , and $c$ from left to right would give the main current direction. And we put each resistor back in between two consecutive nodes, (e.g., $a$ and $b$ , $b$ and $c$ , etc.) according to the two labels nearest to its two ends:

Circuit0013Sol002

And then the simplification is done.

February 19, 2019January 17, 2023

201902190815 Electric circuit diagrams (Elementary) Q12

The blogger claims no originality of his problem below.

Find the equivalent resistance between nodes $a$ and $b$ .

Circuit0003

Solution.

We notice that some resistors will be short-circuited, as shown below:

Circuit0003Sol001

Then, the circuit will have equivalent resistance

$R_{\mathrm{eq}}=\bigg( \displaystyle{\frac{1}{R}+\frac{1}{R}}\bigg)^{-1}+\bigg( \displaystyle{\frac{1}{R}+\frac{1}{R}}\bigg)^{-1}=R$ .

February 19, 2019January 17, 2023

201902190810 Electric circuit diagrams (Elementary) Q11

The blogger claims no originality of his question below.

The circuit below is not a short circuit if switches $S_1$ and $S_2$ are:

Circuit0005

1. $S_1$ : open; $S_2$ : open
2. $S_1$ : open; $S_2$ : closed
3. $S_1$ : closed; $S_2$ : open
4. $S_1$ : closed; $S_2$ : closed

Answer. A

February 19, 2019January 17, 2023

201902190351 Electric circuit diagrams (Elementary) Q10

The blogger claims no originality of his question below.

Find the equivalent resistance between nodes $e$ and $f$ .

Circuit0009

Solution:

We notice that the leftmost resistor is short-circuited, and we have the following diagram:

Circuit0009Sol001

Or, more neatly drawn, we have

Circuit0009Sol002

The equivalent resistance is $R_{\mathrm{eq}}=R+\displaystyle{\frac{R\times R}{R+R}}+R=2.5R$ .

February 19, 2019January 17, 2023

201902190342 Electric circuit diagrams (Elementary) Q9

The blogger claims no originality of his question below.

Find the equivalent resistance between nodes $a$ and $b$ .

Circuit0007

Solution.

The equivalent resistance of two $R$ ‘s in series is $2R$ :

Circuit0007Sol001

The equivalent resistance of $R$ and $2R$ in parallel is $R_{\mathrm{eq}}=\displaystyle{\frac{(R)(2R)}{R+2R}}=\displaystyle{\frac{2}{3}R}$ :

Circuit0007Sol002

For $R$ and $\displaystyle{\frac{2R}{3}}$ in series, $R_{\mathrm{eq}}=\displaystyle{\frac{5}{3}R}$ :

Circuit0007Sol003

Finally, the equivalent resistance between nodes $a$ and $b$ is $R_{\mathrm{eq}}=\displaystyle{\frac{R\bigg( \displaystyle{\frac{5}{3}R}\bigg)}{R+\displaystyle{\frac{5}{3}R}}}=\displaystyle{\frac{5}{8}R}$ .

February 19, 2019January 17, 2023

201902190332 Electric circuit diagrams (Elementary) Q8

The blogger claims no originality of his problem below.

Find the equivalent resistance between nodes $p$ and $q$ .

Circuit0008

Solution.

We draw the following diagram:

Circuit0008Sol001

Then the equivalent resistance is $R_{\mathrm{eq}}=R+\bigg( \displaystyle{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}}\bigg)^{-1}+R=\displaystyle{\frac{8}{3}R}$ .

February 19, 2019January 17, 2023

201902190326 Electric circuit diagrams (Elementary) Q7

The blogger claims no originality of his problem below.

Circuit0006

Find the equivalent resistance of the circuit above.

Solution.

We draw the diagram below:

Circuit0006Sol001

For two resistors $R$ in parallel:

$R_{\mathrm{eq}}=\displaystyle{\frac{R}{2}}$ .

For three resistors $R$ in parallel:

$R_{\mathrm{eq}}=\displaystyle{\frac{R}{3}}$ .

Summing over them in series:

$R_{\mathrm{eq}}=\displaystyle{\frac{R}{2}+\frac{R}{3}+\frac{R}{2}}=\displaystyle{\frac{4}{3}R}$ .

February 19, 2019October 18, 2023

201902190310 Electric circuit diagrams (Elementary) Q6

The blogger claims no originality of his question below.

Circuit0002

In the figure above, the resistors are identical. Find the equivalent resistance between nodes:

$a$ and $b$ ;
$c$ and $d$ ;
$b$ and $c$ .

Solution.

We label the nodes by $A$ , $B$ , $C$ , etc. Different nodes with the same electric potential will be labelled with the same label, as shown below:

Assuming resistors are identical, there is no current flowing through the central resistor between two equipotential end-nodes $B$ ‘s. Hence we draw the diagram below:

The equivalent resistance is $R_{\mathrm{eq}}=\displaystyle{\bigg( \frac{1}{R+R}+\frac{1}{R+R}\bigg)^{-1}}=R$ .(Alternatively one can use the $Y$ – $\Delta$ transformation.)
We draw the diagram below:

The equivalent resistance between nodes $c$ and $d$ is given by $\displaystyle{\frac{1}{R_{\mathrm{eq}}}}=\displaystyle{\frac{1}{R+R}}+\displaystyle{\frac{1}{R}}+\displaystyle{\frac{1}{R+R}}$ . Thus $R_{\mathrm{eq}}=\displaystyle{\frac{R}{2}}$ .
By $Y$ – $\Delta$ transformation.