In nuclear magnetic resonance (NMR) experiments (e.g. hospital MRI scanners). an

Question

In nuclear magnetic resonance (NMR) experiments (e.g. hospital MRI scanners). an inductive coil is used to both transmit and detect radio-frequency signals. Consider a realistic sinusoidal voltage source with output impedance Rs = 50ohm driving a realistic inductive coil (see figure below) which has non-zero resistance. What is the impedance of the series combination of R and L? What is the current through the inductor (L) if its series resistance R = 1ohm and its inductance L = 1 muH? (The sinusoidal voltage source has a 1 V amplitude and omega = 2pi times 7 5MHz.) In Problem 3.4. we discovered that the maximum power dissipated in the load occurs when the load impedance is purely resistive and is equal to the output impedance of the source. Determine the equivalent impedance Zeq of the network to the right of the points A and B in the circuit below. What is Zeq if C1 = 0.639pF. C2 = 3.864 pF. R = 1ohm. L = 1 muH and omega = 2pi times 75MHz? For the case considered in part (

Explanation / Answer

(a) Zs = R + j*w*l =>Zs = 1 + j*2*pi*75*10^6*1*10^-6 =>Series of R & L,Zs = 1+ 150*pij ohm = 1 + 471.24j ohm Series of R & L,Zs = 471.24 ohm Znet = 50 + Zs ohm = 51 + 471.24j =>Znet = 51 + 471.24 j ohm =>Znet = 473.99 ohm I = V/Znet = 1 /473.99 = 2.109 milli Ampere Lagging current (Lags voltage by 83.823 deg ) (b)Z1 = R+j*w*L (R & L are in series) Z2 (Z1 & C2 in parallel) = (Z1/(j*w*C2))/(Z1 + (1/j*w*C2) ) => Z2 = {(R + j*w*L)/(1/j*w*C2)}/{ R+j*w*L + (1/j*w*C2} Zab(Z2 & C1 are in series) = Z2 + 1/(j*w*C1) =>Zab = {(R + j*w*L)/(1/j*w*C2)}/{ R+j*w*L + (1/j*w*C2} + 1/(j*w*C1) Putting values Zab = 3870.106 ohm Znet = Zab + 50 ohm = 3870.44 ohm Inet = V/Znet = 1/ 3870.44 = 0.258 milli Ampere Leading current (Leads voltage by 89.24 deg) I(inductor) = Inet*(1/(j*w*C2))/(Z1 +(1/(j*w*C2) ) I

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