Sir in problem 3 ,for DC gain you have opened capacitors ,but in problem 4, for DC gain you have taken the capacitors and neglected the resistance. Explain this one sir
nice question; I guess it's because of the general configuration of them being in series or parallel; Moreover, thinking more intuitively, try comparing it as low freq=after v high time, then it seems to be clear.
In problem 3, Vin = (voltage drop across parallel combination of R and 2C) + (voltage drop across C). Here if we open circuit the capacitors, we see that there is no current through the resistor and hence Vo = Vin. This still satisfies KVL because voltage drop across the parallel combination of R and 2C is 0V. Whereas, in the 4th problem we have Vin = (voltage across C) + (voltage across series connection of R and 2C). Here at DC, we can open circuit the capacitors. This says that current through the resistance R is 0A and hence voltage drop across the resistance R is 0V. But this DOES NOT tell us anything regarding the voltage across the capacitances C and 2C. Recall that the voltage across a capacitor can be a non zero value even if the current through it is 0. Hence at DC, Vin = Voltage across C + Voltage across 2C in the 4th problem. Hence we perform voltage division by considering both the capacitances in the 4th problem.
Hi Sir, Your lectures are so good, at 14:10 I got gain at high frequency as 2/3 but you solved it to 1/3.
Sir in problem 3 ,for DC gain you have opened capacitors ,but in problem 4, for DC gain you have taken the capacitors and neglected the resistance. Explain this one sir
nice question; I guess it's because of the general configuration of them being in series or parallel; Moreover, thinking more intuitively, try comparing it as low freq=after v high time, then it seems to be clear.
In problem 3, Vin = (voltage drop across parallel combination of R and 2C) + (voltage drop across C). Here if we open circuit the capacitors, we see that there is no current through the resistor and hence Vo = Vin. This still satisfies KVL because voltage drop across the parallel combination of R and 2C is 0V. Whereas, in the 4th problem we have Vin = (voltage across C) + (voltage across series connection of R and 2C). Here at DC, we can open circuit the capacitors. This says that current through the resistance R is 0A and hence voltage drop across the resistance R is 0V. But this DOES NOT tell us anything regarding the voltage across the capacitances C and 2C. Recall that the voltage across a capacitor can be a non zero value even if the current through it is 0. Hence at DC, Vin = Voltage across C + Voltage across 2C in the 4th problem. Hence we perform voltage division by considering both the capacitances in the 4th problem.