RLC Circuit 3

Electricity and Magnetism Level 3

The above is an electric circuit, where a resistor, $2$ inductors and $2$ capacitors are connected to an AC power supply. The maximum voltage and the frequency of the AC power supply is $V_0$ and $\displaystyle \frac{1}{2\pi\sqrt{LC}},$ respectively. If the switch S is connected to $a$ , then the maximum value of the current intensity flowing in the circuit is $I_1,$ and the maximum value of the voltage across the resistor is $V_1.$ Which of the following statements is correct when the switch S is connected to $b?$

a) The impedance of the circuit is $R.$

b) The maximum value of the current intensity flowing in the circuit is $I_1.$

c) The maximum value of the voltage across the resistor is $V_0.$

b) only a) and b) c) only a) only

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Kanishkan R
Apr 7, 2014

total ipidence cannot be R sice the reactance of capasitor and inductance have to be added,and pottential accross the resistance never be equal to the value Vmax as the totel pottential distributed accross all the three component,so option a and c can be ignored.now the total impidence be the capasitor and inductance does not change hence the maximum curent,so option b is right

Abhishek Maji
Sep 18, 2014

When switch is connected to a, the net impedance of the circuit becomes (R+j * sqrt(L/C)) and when the switch is connected to b , the net impedance offered by the circuit becomes (R-j * sqrt(L/C)) at that particular supply frequency....So in both the cases the circuit impedance remains same only difference is that in the first case it is Inductive and later it is Capacitive..Therefore for same excitation voltage the maximum value of the current intensity flowing in the circuit and the maximum value of the voltage drop across the resistor remains same in both the cases....therefore statement in b) is correct...

Kumar Satish
Mar 29, 2014

Case A: XL= 2 times Xc i.e root under(L/C) . case B: Xc = 2 times XL (sqrt.L/C) . so net impedance R+(XL-Xc)j dont change and hence maximum value of current is same (i1)

Tarashankar Bhattacharya
Mar 28, 2014

when connected to a: total resistance of circuit R(total)=R+j(2wl-(1/wc)) where w=2 pi frequency=1/(LC)^1/2 so current is i1=v0/R(total) taking real part only and putting the value of w i1=v0R/(R^2-L/C)

When point b is connected: total resistance of circuit R(total)=R+j(wl-(2/wc)) By taking real part only we get i2=i1 so both the current are same is only right option.

I may be wrong but... On the EXACTLY moment when you turn switch to b the impedance will be only R, as the capacitor and inductor will work as a wire. As the time passes, they will charge and modify the impedance. So a) should be correct...

Considering only the real part of the current dramatically changes the actual current. When you are talking of the current that flows the circuit, it should be the MODULUS of the complex number that represents the impedance. Actually you can ONLY take only the real part if you consider that a) is correct, so on this exact moment the current will be V/R.

So we have 2 options: If you consider that a) is wrong (it's not the exact moment that you turn the switch), then necessarily "b" is also wrong as stated, If you consider that a) is right, then b) is also right.

Even if I am wrong and on the exactly moment that you switch to b the impedance isn't only R, b) should be wrong as the current will NOT be only the real part, making the question have no answer.

Felipe Magalhães - 7 years, 2 months ago

Why not c

Kahsay Merkeb - 7 years, 2 months ago

i also thought the same

zalmey khan - 7 years, 1 month ago

RLC Circuit 3

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