3-16the tank current is in phase with the primary voltage e_{p}. The current flowing in the tank circuit causesvoltages e_{1} and e_{2} to be developed in the secondary winding of T1. These voltages are of equal magnitudeand of opposite polarity with respect to the center tap of the winding. Since the winding is inductive, thevoltage drop across it is 90 degrees out of phase with the current through it.Figure 3-13 is a simplified schematic diagram of a ratio detector at resonance. The voltage applied tothe cathode of CR1 is the vector sum of e_{1} and e_{p}. Likewise, the voltage applied to the anode of CR2 isthe vector sum of e_{2} and e_{p}. No phase shift occurs at resonance and both voltages are equal. Both diodesconduct equally. This equal current flow causes the same voltage drop across both R1 and R2. C3 and C4will charge to equal voltages with opposite polarities. Let’s assume that the voltages across C3 and C4 areequal in amplitude (5 volts) and of opposite polarity and the total charge across C5 is 10 volts. R1 and R2will each have 5 volts dropped across them because they are of equal values. The output is taken betweenpoints A and B. To find the output voltage, you algebraically add the voltages between points A and B(loop ACB or ADB). Point A to point D is -5 volts. Point D to point B is + 5 volts. Their algebraic sum is0 volts and the output voltage is 0 at resonance. If the voltages on branch ACB were figured, the sameoutput would be found because the circuit branches are in parallel.Figure 3-13.—Current flow and polarities at resonance.When the input signal reverses polarity, the secondary voltage across L2 also reverses. The diodeswill be reverse biased and no current will flow. Meanwhile, C5 retains most of its charge because of thelong time constant offered in combination with R1 and R2. This slow discharge helps to maintain theoutput.OPERATION ABOVE RESONANCE.—When a tuned circuit (figure 3-14) operates at afrequency higher than resonance, the tank is inductive. The secondary current ilags the primary voltagee_{p}. Secondary voltage e_{1} is nearer in phase with primary voltage e, while e_{2} is shifted further out of phasewith e_{p}. The vector sum of e_{1} and e_{p} is larger than that of e_{2} and e_{p}. Therefore, the voltage applied to thecathode of CR1 is greater than the voltage applied to the anode of CR2 above resonance.Figure 3-14.—Current flow and polarities above resonance.