By A. M. P. Brookes and P. Hammond (Auth.)
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Extra info for Advanced Electric Circuits
1 shows two voltage generators driving current round a circuit. 2s/2 cos cut T Amps I Amps FIG. 1 (a) Express the voltage excitation in complex form. (b) Set up the circuit equation. 32 ADVANCED ELECTRIC CIRCUITS (c) Obtain the modulus and phase angle of /, and then turn the resulting complex quantity back into a cosine. ^/is:(a)3e^, [V(3)-j]e^, ,3-V(3)+j (b) / = QW x' l+j (c) modulus 1-14, phase angle -6-7° ( =38-3° -45°) M4cos(ft>/-6-7°)A 2. 2 shows two current generators feeding a load. Carry out steps similar to (a), (b) and (c) in Question 1, for the node potential.
The circuit operates efficiently only if the additional current through Z causes a large voltage drop across Ro and thus reduces the current through R at the unwanted frequency. For this to happen, Z must be small compared with RQ. We have seen that the circuit arrangement of Fig. 3(a) operates satisfactorily if R and Ro are both fairly low while that of Fig. 3(b) works well when R and Ro are both fairly large. Cases arise, however, where R may be large and Ro small, or vice versa. When this happens, the addition of a single impedance Z cannot normally provide satisfactory filtering, so we must consider the addition of two impedances.
Example: The bridge shown in Fig. 5 is balanced at a frequency of 2-5 kc/s when C\ has a pure capacitance of 1 /xF. Find the capacitance and power factor of the capacitor C2. 1000& 800X1 FIG. -. -. -. power factor = cos^ - ro»C2 = 0-0314 Some bridges involve self and mutual inductance as shown in the next example of a frequency-measuring bridge. d. applied to terminals A and B. AO 30mHc w s >30mH > 20 XI 10 mH 1 Detector 200 a J ±c BO T F I G . 6 The ideal capacitance C is fixed at 2 /xF and the mutual inductance M between the coils is variable from 0 to 20 mH.