At resonance, the L and C impedance cancels out, so the current can be calculated based on the resistance and applied voltage.
Imagine increasing frequency of the supply from 0 Hz to very high. At low frequency, the impedance of the inductor is ~0 (defined as Zl = w*L*j), and the impedance of the capacitor is very large (defined as Zc = 1 / (w*C*j)). As you increase the frequency, the impedance of the capacitor will decrease, as the impedance of the inductor increases. At some point (the resonant frequency), these two will be equal, with opposite signs. After crossing the resonant frequency, the inductor impedance will continue growing larger than the capacitor impedance until the total impedance approaches infinite.
When the frequency of Parallel RL Circuit Increases,XL increases which causes IL (current through inductor) decreases. Decrease in IL causes It (It=Il+Ir) to decrease,which means by relation IT=Vs/Zt ,the Zt (Total Impedance) Increases.
Increases the total resistance
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
Inductive reactance case of ac) is equivalent to resistance (in case of dc) for inductors.So if resistance increases current decreasesas well as if inductive reactance increases current decreases
Then the brightness of the light buld increases.
When the frequency of Parallel RL Circuit Increases,XL increases which causes IL (current through inductor) decreases. Decrease in IL causes It (It=Il+Ir) to decrease,which means by relation IT=Vs/Zt ,the Zt (Total Impedance) Increases.
Increases the total resistance
A capacitor stores accumulated charges as voltage. The storing and discharging of these charges help regulate voltage in circuits. As the switching frequency of storing or discharging increases the 'frequency dependant resistance' or reactance decreases and appears as a short circuit at infinite frequency. The capacitor appears as a open circuit at zero frequency.
Current increases if the voltage remains constant.
As the number of bulbs in a series circuit increases, the current decreases. As the number of bulbs in a parallel circuit increases, the current increases.
Increases
current decreases and resistance increases
when the frequency is increased the total impedance of a series RC circuit is decrease.
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
No, the total resistance increases.
Inductive reactance case of ac) is equivalent to resistance (in case of dc) for inductors.So if resistance increases current decreasesas well as if inductive reactance increases current decreases
Then the brightness of the light buld increases.