Kinda tricky item in a question about an electrical circuit

[ksio89]

I'm trying to solve item "c" of this question about the electrical circuit depicted on the attached picture.

The question is in Brazilian Portuguese, this would be a rough and not 100% gramatically correct translation:

"In the circuit shown below on Pic. 1 (labeled "Electrical Circuit"), the three resistors have values $R_1$ = 2 Ω, $R_2$ = 20 Ω and $R_3$ = 5 Ω. The battery B has constant tension 12 V. The current $i_1$ is considered positive in the indicated direction. Between the instants $t$ = 0 s and $t$ = 100 s, the generator G provides a variable tension $V = 0.5 t$ (V in volts and t in seconds).



a) Define the current value $i_1$ for $t$ = 0 s

b) Define the instant $t_0$ in which the current $i_1$ is null.

c) Plot the curve which depicts the current $i_1$ in function of the time $t$, in the interval from 0 to 100 s. Use the axis from the following picture (labeled "Graph") and clearly indicate the scale of current, in ampère (A).

d) Define the value of the power $P$ received or provided by the battery $B$ in the instant $t$ = 90 s."


I was able to solve items a, b and d using Kirchoff's Voltage Law and generating a linear system with two variables, $i'_1$ and $i'_2$, where $i'_1 \pm i'_2 = i_1$ and $i'_1$ is the current flowing around the left loop (loop I) and $i'_2$ is the current flowing around the right one (loop II). So:

a) 2 s;

b) $t$ = 30 s (in this case, $i'_1 = -i'_2$;

c) ?

d)$P$ = (I need to find the graph of $i_1$ vs. $t$ so I can calculate the absolute value of the power).

I wasn't able able to answer item c, because it asks to plot the graph of $i_1$ Vs $t$. The issue is that $i_1$ is actually the algebric sum of $i'_1$ and $i'_2$, but I also have a third variable which is the tension $V$ generated by the voltage source G, which is a function of time $t$. I don't have a third equation though, as Kirchoff's Voltage Law applied to the largest loop provides me an equation which is simply a linear combination of the equations associated with loops I and II.

According to the book answers section, the answer for item c) is a line of slope-intercept form $f(t)= -\frac{t}{15}+2$. Then, |$P$| would be 48 W. But how do you define $i_1$ as a function of $V$ and, consequently, $t$?

 

[haruspex]

Just fill in unknowns for voltages along the various wires. There are three of these. Also an unknown for the current through each resistor.
You should get an equation relating these to the drop across each resistor and two more for the drops across the voltage sources. Finally, an equation relating the three currents.
Eliminating the extra unknowns should produce an equation for i1.

 

[NascentOxygen]

I wasn't able able to answer item c, because it asks to plot the graph of i1 Vs t. The issue is that i1 is actually the algebric sum of i′1 and i′2, but I also have a third variable which is the tension V generated by the voltage source G, which is a function of time t. I don't have a third equation though, as Kirchoff's Voltage Law applied to the largest loop provides me an equation which is simply a linear combination of the equations associated with loops I and II.

You have the loop currents, which I'll denote i_a and i_b. Mark as v_a the voltage at the junction of the 3 resistors. You can write v_a in terms of R₂ and its current. Also, write v_a in terms of R₃ and G. You can also get another equation with the current through R_1.

You are not interested in v_a so eliminate it from those equations, keeping as a single term (i₁ + i₂) because that's the independent variable you are needing for your graph.

 

[ksio89]

Thanks for the replies, but I still can't solve the question. I have four variables, $i_1$, $i_a$, $i_b$ and $V$, but only three independent linear equations.

Let's say $i_1$ = $i_a+i_b$, where $i_a$ is the current on left loop which flows counterclockwise, and $i_b$ on the right one, clockwise. The current $i_a$ flows upwards.

Equation 1: from point A (on the junction of 3 resistors) to point A, around the left loop, clockwise:

+$R_1i_1$ - 12 + $V$ + $R_3i_a$ = 0


Equation 2: from point A to A, around the right loop, clockwise:

-$R_2i_b$ + 12 -$R_1i_1$ = 0

adding the two equations gives:

-20$i_b$ + $V$ + 5$i_a$ = 0

$V$ = 20$i_b$ - 5$i_a$.

I didn't even need to add the emf sum/drops of the section with the battery and resistor $R_1$, because each loop has it in common, but with opposite signs, which is equivalent meaning it's KVL applied to the big loop beginning and ending on A, clockwise.

The issue is that besides $V$ = 20$i_a$ - 5$i_b$, the only other equation I have is $i_1$ = $i_a+i_b$, which is not enough to solve this system of equations.

Sorry, but I really don't know where to go now. I simply can't see how you define $i_1$ as function of $i_a$ and $i_b$. I would be grateful if someone showed me the remaining equations and instructed on how they were generated.

 

[gneill]

The only variables you have for your mesh analysis are the two mesh currents, $i_a$ and $i_b$. $i_1$ is a derived value that you will later calculate from $i_a$ and $i_b$, and V will take on values that come from its definition (V = 0.5t). For now treat V as a constant and lug it around in the equations as such.

So, standard mesh analysis to find $i_a$ and $i_b$. You will end up with expressions for these currents involving V.

 

[gneill]

Alternatively, use bare-bones KCL and KVL. Define the other two currents $i_2$ and $i_3$ and write KVL for the two loops and KCL at the node:

You'll have three equations in the current unknowns. Use elimination to solve for $i_1$.

 

[NascentOxygen]

You have the loop currents, which I'll denote i_a and i_b. Mark as v_a the voltage at the junction of the 3 resistors. You can write v_a in terms of R₂ and its current. Also, write v_a in terms of R₃ and G. You can also get another equation with the current through R₁, viz., (i_a+i_b).

You are not interested in v_a so eliminate it from those equations, keeping as a single term (i_a + i_b) because that's the independent variable you are needing for your graph.

Correction incorporated.