1.8 k2O A.-200 C2 | 10F 100 F 8 V. 1.0 kHz 100 Figure 3 is approximately 20) Ref
ID: 2249955 • Letter: 1
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1.8 k2O A.-200 C2 | 10F 100 F 8 V. 1.0 kHz 100 Figure 3 is approximately 20) Refer to Figure A) 25v B)75 V Csov D) 10V Rin is approximately B) 1.0k 21) Refer to Figure 21) A) 27k0 91.8 D)4542 22) The amplifier type that frequently has a resonant circuit load is A) class-C B) class-AB dlass-B D) class-A 23) The critical frequency of a high-pass filter is where the response is 23) A) at its highest point C)-20 dB from the passhand response B) -3 dB from the passband response D)-6dB from the passband response R150 kn 22 10 27 Figure 2 24) Refer to Figure 2(a). The closed-loop gain, ACL, is 24) A)-78 B) +78 C)-6.8 D) +68 25) Refer to Figure 2(c). The closed-loop gain, ACL, is B) 9.2 C)-92 D)+82Explanation / Answer
Answer:-20) The value of Thevenin's voltage is-
ETh = (R2 x Vcc )/(R1 + R2 ) = 9 V. Thevenin's resistance RTh = (R1 x R2)/(R1 + R2) = 1.08 kOhm. Hence the base current IB = (ETh - VBE)/(RTh + (beta + 1)*RE) = 391.87 micro-A. Hence emitter current IE = (beta + 1)*IB = 78.76 mA. So VE = IE x RE = 7.87 V. So option B is correct.
Answer:-21) re = 0.026/IE = 0.33 ohm. Hence Zb = (beta*re) + (beta + 1)*RE = 20.166 kOhm. So input impedance Zin = R1 || R2 || Zb = 1.025 kOhm. Hence option B is correct.
Answer:-23) Option B is correct. Critical frequency or cut-off frequency is the point where gain is -3 dB from passband.
Answer:-24) Option C is correct. For inverting amplifier gain is G = -(Rf / R1) = -150/22 = -6.81.
Answer:-25) Option B is correct. For non-inverting amplifier gain G = (1 + (Rf / R1) ) = 1 + (82/10) = 9.2.
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