Creating two branches with similar current using capacitors/resistors

Question

I am studying for my midterm practical and I am stuck on a basic question. Below is a circuit that will help explain (made in Multisim).

So, we are given this basic structure. All components must stay in their given locations, but their values may change ,but the currents in each individual branch comes out dramatically different when using a Single Frequency Analysis. I assumed C1 = 1.7uF and C2 = 4.7uF because my professor told me that I needed to make assumptions for some values.

My work:

X represents impedance.

$$X_{C_1}= \frac1{2πfC_1}=\frac1{2*π*60*1.7*10^{-6} }=1560.34Ω $$

$$X_{C_2}= \frac{1}{2πfC_2 }=\frac1{2*π*60*4.7*10^{-6}}=564.38Ω $$

$$X_{C_1} = X_{C_2} + R_2 $$ $$ R_2 = X_{C_1} - X_{C_2}$$ $$ R_2 = 995.96Ω$$

However, the circuit demonstrates a minimal current difference when R2 is around 1500Ω. What factors am I not including?

Answer 1

\$X\$ represents reactance not impedance. Impedance offered by capacitance is \$-jX_C\$ and it is a complex value.

So the impedance offered by branch-1 is \$-jX_{C_1}\$ and that offered by branch-2 is \$-jX_{C_2} + R_2\$

For equal current to flow through these branches, the magnitude of impedances should be equal. ie.,

$$|-jX_{C_1}| = |-jX_{C_2} + R_2|$$ $$R_2 = \sqrt{X^2_{C_1} - X^2_{C_2}}$$ $$R_2 = \sqrt{1560.34^2-564.38^2} = 1454.69\Omega$$

You missed the \$-j\$ and wrote \$X_{C_1} = X_{C_2} + R_2\$ . That made the calculations go wrong.