February 19, 2019 – Physicspupil

February 19, 2019January 17, 2023

201902190842 Electric field representation

The blogger claims no originality of his question posted here.

Think of a field in some representation other than a diagram of field lines. For instance, suppose any point in a field is assigned a number to its field strength magnitude, but the direction of field strength of each point is not indicated. Now, could we deduce a rough picture of the (vector) E-field from these (scalar) numbers, provided that we know the distribution of these numbers and the position of charges?

Visualizing_the_Efield_by_numbers_distribution

In the figure, $A$ , $B$ and $C$ are charges of unknown charge quantities and sign. Each single-digit number at a point represents the magnitude of E-field strength at that point. For example, $6$ stands for $6\,\mathrm{N/C}$ , $0$ stands for zero E-field, i.e., a neutral point. By deducing from the figure, determine which of the following statements is/are correct.

(1) Charge quantity of $A$ is larger than that of either $B$ or $C$ .
(2) Charge quantity of $B$ equals charge quantity of $C$ .
(3) Charge $A$ is positive. Charge $B$ and charge $C$ are both negative.

A. (1) only
B. (2) only
C. (1) and (2) only
D. (1), (2), and (3)

Answer. D

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.

February 19, 2019January 17, 2023

201902190246 Electric circuit diagrams (Elementary) Q5

The blogger claims no originality of his problem below.

Circuit0012

Find the equivalent resistance of the circuit above.

Solution.

In series, $5\,\mathrm{\Omega} +9\,\mathrm{\Omega} +1\,\mathrm{\Omega} =15\,\mathrm{\Omega}$ .

Circuit0012Sol001

In parallel, $\bigg(\displaystyle{\frac{1}{10\,\mathrm{\Omega} }+\frac{1}{15\,\mathrm{\Omega} }}\bigg)^{-1}=6\,\mathrm{\Omega}$ .

Circuit0012Sol002

In series, $4\,\mathrm{\Omega} +6\,\mathrm{\Omega} +2\,\mathrm{\Omega} =12\,\mathrm{\Omega}$ .

Circuit0012Sol003

In parallel, $\bigg(\displaystyle{\frac{1}{4\,\mathrm{\Omega} }+\frac{1}{12\,\mathrm{\Omega} }}\bigg)^{-1}=3\,\mathrm{\Omega}$ .

Circuit0012Sol004

Finally, the equivalent resistance of the resistors above in series is $R_{\mathrm{eq}}=3\,\mathrm{\Omega} +3\,\mathrm{\Omega} +4\,\mathrm{\Omega} =10\,\mathrm{\Omega}$ .