4
\$\begingroup\$

enter image description here

My goal is to calculate the current i(t) up in the right corner. Using the formula $$ Z_{total} = \left( U\over I\right), I = \left( U\over Z_{total}\right) $$

Given values are

$$ R_{1} = 1k \Omega,\ R_{2} = 0.5k \Omega,\ R_{3} = 0.4k \Omega,\ C_{1} = 1 \mu F,\\ C_{2} = 1 \mu F,\ L_{1} = 2H,\ e(t) = 10sin(1000t),\ \omega = 1000 $$ The internal impedance of the source is $$ Z_{i} = 10e^{j\pi/4}k \Omega $$

I start with the parallell connection between $$ Z_{R2}//Z_{C1} = \left( Z_{R2}*Z_{C1}\over Z_{R2}+Z_{C1}\right) = \ \left( 500*\left( 1\over 0.001j\right) \over500+\left( 1\over 0.001j\right)\right) = -400 + 200j $$

Ztotal: $$ Z_{total} = Z_{R1} +Z_{R2}//Z_{C1}+Z_{C2} + Z_{R3} +Z_{L1} + Z_{i} $$ Im not sure if the internal impedace of the source Z(i) is supposed to be added to the total impedance as if it is connected in series with the rest of the impedances? If it is I get the following.

$$ Z_{total} = 1000 +(-400 + 200j) -1000j + 400 + 2000j + 10\times1000(\cos(\pi/4)+j\sin(\pi/4)$$ $$= (1000+5000\sqrt2)+j(1200+5000\sqrt2) $$

Now I use the formula above, since the current goes from + to - I get a negative voltage. $$ I = \left( -U\over Z_{total}\right) = \left( -10\over (1000+5000\sqrt2)+j(1200+5000\sqrt2)\right) $$ $$=0.000865e^{-j0.007976} $$ $$= 0.000865\sin(1000t-0.007976) $$ Does this make any sense?

\$\endgroup\$

1 Answer 1

2
\$\begingroup\$

What ever you have done is correct except at the last step.

The current flows from +ve terminal to -ve terminal and the current \$i(t)\$ marked in the schematic is in the same direction. So the voltage you have to consider is +U. That will change the phase angle by \$\pi\$. ie., $$I = \left( U\over Z_{total}\right) = \left( 10\over (1000+5000\sqrt2)+j(1200+5000\sqrt2)\right)$$ $$ = 0.000865e^{-j(0.007976+\pi)}$$ $$= 0.000865\sin(1000t-0.007976-\pi)$$

The ac current through the circuit is going to be a sinusoid with same frequency as the voltage source, phase shifted (lag) by \$\mathbf{0.007976+\pi}\$ and peak value of \$0.865mA\$.

Since there are reactive elements in the circuit, the current will have a phase difference from the voltage. So this makes perfect sense.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.