The tables below shows what to submit formulas or equations using the default 1-D Equation Editor in CircuitsU. If you want to submit multiple equations in the same textbox you needs to write each equation on a separate line.
To submit a single line answer (such as the final answer in a problem) you can type the final numerical value with or without any units or the analytical formula. Notice that
CircuitsU also has a 2-D Equation Editor that is easier to use and can be activated by going to Manage your account > Settings.
The 2-D Equation Editor is currently implemented only in a few type of problems.
Table 1. How to submit numerical answers.
| To submit... |
type... |
| $1.234\times10^{-3}$ | 0.001234
1.234e-3
1.234m (use postfixes for powers of 10: n for $10^{-9}$, u for $10^{-6}$, m for $10^{-3}$, k for $10^3$, M for $10^6$, and G for $10^9$) |
| $I_0=2\:{\textcolor{gray}A}$ | I0=2 A
I0=2 (units are optional)
2 A (you don't need to specify the sought variable if you submit a single-line answer)
2 |
| $V_0=2\:{\textcolor{gray}{kV}}$ | V0=2000 V
V0=2000 (units are optional)
2000
2e3
2k |
| $45°$ | 45 deg
45deg
45° |
| $45 \: \textcolor{gray}{rad}$ | 45 rad
45rad |
Table 2. How to submit mathematical formulas using the 1-D Equation Editor.
| To submit... |
type... |
| $R_1+R_2+R_3$ | R1+R2+R2 |
| $\frac{R_1 R_2}{R_1+R_2}$ | R1*R2/(R1+R2) (the multiplication sign between R1 and R2 is mandatory in this case) |
| $\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}}$ | 1/(1/C1+1/C2+1/C3) or (1/C1+1/C2+1/C3)^(-1) |
| $\frac{1}{\textcolor{blue}{j}\omega C}+j\omega L$ | 1/(jwC)+jwL |
| $P_d= R_1 (i_2-i_3)^2$ | Pd=R1*(i2-i3)^2
Pd=R1(i2-i3)^2 (the multiplication sign is optional when multiplying a number with a symbolic quantity)
R1(i2-i3)^2 (it is optional to specify the sought variable if you submit a single-line answer) |
| $2\times 10^6\times [3-(a+b)]$ | 2M*(3-(a+b)) (replace square brackets with paranthesis)
2M(3-(a+b)) |
| $2 i_1+ R_1 I_x -7 = 0$ | 2*i1+R1/Ix-7=0 (if you submit an equation make sure you don't forget the equal sign)
2i1+R1/Ix-7=0 |
| $i_1=\frac{v_1-v_2}{R1}-\frac{v_3}{6.5}$ | (v1-v2)/R1-v3/6.5=0 |
| $\frac{i_1}{2 \textcolor{blue}{j}}+3j(i_1-i_2)$ | 0=i1/(2*j)+3*j*(i1-i2) ($\textcolor{blue}{j}$ denotes imaginary number $\textcolor{blue}{j}=\sqrt{-1}$)
0=i1/(2j)+3j(i1-i2) |
Table 3. How to submit differential and integral equations using the 1-D Equation Editor.
| To submit... |
type... |
| $\dfrac{dv_{1}(t)}{dt}$ | derivative(v1(t),t) |
| $\dfrac{dv_{1}}{dt}$ (incorrect) | It is incorrect to use derivative(v1,t) (see the above line for the correct expression). You need to show the time dependance explicitly when referring to time-dependent quantities in CircuitsU. Therefore, you need v1(t) instead of v1. |
| $\int_{0}^{t}i_{1}(t){dt}$ | integral(i1(t),t) |
| $\int_{0}^{t}i_{1}{dt}$ (incorrect) | It is incorrect to use integral(v1,t) (see the above line for the correct expression). You need to show the time dependance explicitly when referring to time-dependent quantities in CircuitsU. Therefore, you need i1(t) instead of i1. |
| $I(t)=C\dfrac{d[v_{1}(t)-v_{2}(t)]}{dt}$ | I(t)=C*derivative(v1(t)-v2(t),t) |
| $I(t)=\frac{1}{L}\int_{0}^{t}[v_{1}(t)-v_{2}(t)]{dt}$ | I(t)=1/L*integral(v1(t)-v2(t),t) |
Note that the constants and functions shown in see Table 4 might not always be available when you submit answers. Their availability depends on what you are asked to submit at that particular step.
Table 4. Predefined constants and functions.
| Constant/Function |
Code |
| $\textcolor{blue}{j}=\sqrt{-1}$ | j |
| $\pi$ | pi or 3.14 |
| $e^x$ | exp(x)
e^x |
| $a^b$ | a^b |
| $\sqrt{x}$ | sqrt(x)
x^0.5 |
| $\sin(x)$ | sin(x) |
| $\cos(x)$ | cos(x)
E.g.: cos(45 deg) or cos(0.785 rad) |