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The frequency would decrease the gain and increase the output voltage
That depends on the circuit. For a pure resistive circuit (no inductance and capacitance), the frequency will have no effect on the current.
coupling capacitors are generally used to couple the the AC component of voltage to the DC component(biased voltage) of the transistor amplifier . As we know that the capacitor itself has some reactance which is variable with the applied frequency Rc=1/wc where w=frequency in radians = 2*pi*f and f= frequency of circuit. and, V=VC+VIN VC= voltage drop on capacitor VIN= resultant voltage available for the transistor for amplification so as, frequency increases reactance decreases drop on C decreases so, voltage available for transistor increases and now you can analyse yourself for the case if frequency decreases
The effect of diode voltage drop as the output voltage is that the input voltage will not be totally transferred to the output because power loss in the diode . The output voltage will then be given by: vout=(vin)-(the diode voltage drop).
voltage would go up typically (you would raise the voltage, not the fan circuit), but it depends on the motor
The frequency would decrease the gain and increase the output voltage
A: Those diodes are used as a variable capacitance whereby applying a voltage changes the capacitance which effect the frequency of the tuning circuit
The effect of an RL circuit in half wave rectifier is that the voltage output wave forms for current and voltage will be modified .
That depends on the circuit. For a pure resistive circuit (no inductance and capacitance), the frequency will have no effect on the current.
There will be no effect on the voltage. That is the effective voltage will be only 12 volt. But there will be increase of current.
coupling capacitors are generally used to couple the the AC component of voltage to the DC component(biased voltage) of the transistor amplifier . As we know that the capacitor itself has some reactance which is variable with the applied frequency Rc=1/wc where w=frequency in radians = 2*pi*f and f= frequency of circuit. and, V=VC+VIN VC= voltage drop on capacitor VIN= resultant voltage available for the transistor for amplification so as, frequency increases reactance decreases drop on C decreases so, voltage available for transistor increases and now you can analyse yourself for the case if frequency decreases
The effect of diode voltage drop as the output voltage is that the input voltage will not be totally transferred to the output because power loss in the diode . The output voltage will then be given by: vout=(vin)-(the diode voltage drop).
voltage would go up typically (you would raise the voltage, not the fan circuit), but it depends on the motor
A parallel circuit is different in many ways from a series circuit: 1. In parallel, the voltage across all the devices connected is the same. 2. If a fault occurs in any device connected in parallel combo, then it has no effect on the operation of the other device. 3. In series circuit the current flowing through all the devices is the same while in case of the parallel one the voltage across all the devices is same.
The input impedance should increase slightly for the lower frequency, when using a capacitive circuit.
it will cause a Short Circuit
Ohm's Law answers your question. Voltage = Current x Resistance. In a series circuit you are in effect adding resistance. If the Voltage remains constant then the answer is obvious looking at the equation above.