Capacitor and inductor in parallel

  • #1
Titan97
Gold Member
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Homework Statement


[/B]
Find the power supplied by the battery.

Capture.PNG


Homework Equations


$$Z_C=-iX_C$$
$$Z_L=iX_L$$

The Attempt at a Solution


$$Z_1=R_1-iX_C=\sqrt{R_1^2+X_C^2}e^{i\phi_1}$$
$$Z_2=R_+iX_L=\sqrt{R_2^2+X_L^2}e^{i\phi_2}$$
$$I_1(t)=\frac{V_0}{\sqrt{R_1^2+X_C^2}}e^{i(\omega t-\phi_1)}$$
$$I_2(t)=\frac{V_0}{\sqrt{R_2^2+X_L^2}}e^{i(\omega t-\phi_2)}$$

##\phi_1=\arctan{\frac{12}{5}}##
##\phi_2=\arctan{\frac{4}{3}}##

$$i_{1,rms}=\frac{V_{rms}}{\sqrt{R_1^2+X_C^2}}=10A$$
$$i_{2,rms}=\frac{V_{rms}}{\sqrt{R_2^2+X_L^2}}=13A$$

Power dissipated is ##23\times 130\text{W}##

But the answer given is ##1514\text{W}##
 

Answers and Replies

  • #2
cnh1995
Homework Helper
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It looks like the problem asks for active power.The active power comes out to be 1513.4W(close to the given answer). Start by finding the expression for current in complex form.
ϕ1=arctan125ϕ1=arctan⁡125\phi_1=\arctan{\frac{12}{5}}
This should be arctan (-12/5). The impedance in the capacitive branch is 5-j12 ohm.
 
  • #3
ehild
Homework Helper
15,543
1,912

Homework Statement


[/B]
Find the power supplied by the battery.

View attachment 100745

Homework Equations


$$Z_C=-iX_C$$
$$Z_L=iX_L$$

The Attempt at a Solution


$$Z_1=R_1-iX_C=\sqrt{R_1^2+X_C^2}e^{i\phi_1}$$
$$Z_2=R_+iX_L=\sqrt{R_2^2+X_L^2}e^{i\phi_2}$$
$$I_1(t)=\frac{V_0}{\sqrt{R_1^2+X_C^2}}e^{i(\omega t-\phi_1)}$$
$$I_2(t)=\frac{V_0}{\sqrt{R_2^2+X_L^2}}e^{i(\omega t-\phi_2)}$$

##\phi_1=\arctan{\frac{12}{5}}##
##\phi_2=\arctan{\frac{4}{3}}##

$$i_{1,rms}=\frac{V_{rms}}{\sqrt{R_1^2+X_C^2}}=10A$$
$$i_{2,rms}=\frac{V_{rms}}{\sqrt{R_2^2+X_L^2}}=13A$$


Power dissipated is ##23\times 130\text{W}##

But the answer given is ##1514\text{W}##
The rms value of sum of complex currents is not equal the sum of the rms values. You know that only resistors dissipate power. The power dissipated in both parallel circuits is (Irms)2*R. Calculate (i1rms)2R1+(i2rms)2R2
 
  • #6
Titan97
Gold Member
451
18
But why:
The rms value of sum of complex currents is not equal the sum of the rms values
 
  • #7
ehild
Homework Helper
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AC currents and voltages can be represented by their complex amplitudes-by complex numbers. The rms value is (1/√2) times the modulus of the complex voltage and current.
If you add two complex numbers z1 = a+jb and z2 = c+jd,
z1+z2 = (a+c)+j(b+d),
and the modulus is
##|z1+z2|=\sqrt{(a+c)^2+(b+d)^2} ##
which is not equal the sum of the absolute values
##|z1|=\sqrt {a^2+b^2}## and |##z2|=\sqrt {c^2+d^2}##.
As the rms value is 1/√2 times the magnitude, the rms values do not add, either.

As an example: If you have a capacitor of impedance -10j Ω and an inductor of impedance 10j Ω parallel to an AC source of rms voltage 100 V, the rms currents through both the capacitor and inductor are 10 A, but the total current is zero, as the complex currents cancel. I=U/(-10)+U/10) = 0.
 
  • #8
cnh1995
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Gold Member
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But why:
RMS values are computed over a cycle. Here, rms values of currents are 10A and 13A which are time independent, because they are calculated over a complete cycle. If they were instantaneous currents, they could be added directly since both appear at the same time.
 

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