When a capacitor is connected in series LR circuit, the alternating current flowing in the circuit increases. Explain why.
Download our app for free and get started
Impedance of series LR circuit
$\text{Z}_1=\sqrt{\text{R}^2+\text{X}^2_{\text{L}}}$
When capacitor is also connected in circuit,
Then impedance
$\text{Z}_{\text{L}}=\sqrt{\text{R}^2+(\text{X}_{\text{L}}-\text{X}_{\text{C}})^2}$
Clearly impedance of circuit decreases $(Z_2 < Z_1)$, so the value of current $\text{I}=\frac{\text{v}}{\text{z}}\propto\frac{1}{\text{z}}$ in the circuit increases.
$\text{Z}_1=\sqrt{\text{R}^2+\text{X}^2_{\text{L}}}$
When capacitor is also connected in circuit,
Then impedance
$\text{Z}_{\text{L}}=\sqrt{\text{R}^2+(\text{X}_{\text{L}}-\text{X}_{\text{C}})^2}$
Clearly impedance of circuit decreases $(Z_2 < Z_1)$, so the value of current $\text{I}=\frac{\text{v}}{\text{z}}\propto\frac{1}{\text{z}}$ in the circuit increases.
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A source of ac voltage $ v= v_0 \sin\omega t$, is connected across a pure inductor of inductance L. Derive the expressions for the instantaneous current in the circuit. Show that average power dissipated in the circuit is zero.View Solution
- 2View SolutionState Ampere's circuital law. Use this law to find magnetic field due to straight infinite current carrying wire. How are the magnetic field lines different from the electrostatic field lines?
- 3An AC source of voltage $\text{V = Vm}\sin\omega\text{t}$ is applied across a series LCR circuit. Draw the phasor diagrams for this circuit, when,View Solution
- Capacitive impedance exceeds the inductive impedance.
- Inductive impedance exceeds capacitive impedance.
- 4Two harmonic waves of monochromatic light$\text{y}_{1} = \text{a} \cos\omega\text{t}\text{ and } \text{y}_{2} = \text{a} \cos(\omega\text{t} + \Phi)$View Solution
are superimposed on each other. Show that maximum intensity in interference pattern is four times the intensity due to each slit. Hence write the conditions for constructive and destructive interference in terms of the phase angle$\Phi$. - 5View SolutionConsider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.
- 6View SolutionDraw the effective equivalent circuit of the circuit shown in Fig, at very high frequencies and find the effective impedance.
- 7View Solution
- When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
- A lamp is connected in series with an inductor and an AC source. What happens to the brightness if the lamp when the key is plugged in and an iron rod is inserted inside the inductor? Explain.
- 8View SolutionA 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 300 pF capacitor. Calculate how much electrostatic energy is lost in the process. What is the source of energy loss?
- 9Two capacitors of capacitance $10\mu\text{F}$ and $20 \mu\text{F}$ are connected in series with a 6 V battery. After the capacitors are fully charged, a slab of dielectric constant (K) is inserted between the plates of the two capacitors. How will the following be affected after the slab is introduced:View Solution
- The electric field energy stored in the capacitors.
- The charges on the two capacitors.
- The potential difference between the plates of the capacitors.
- 10In the given circuit, the value of resistance effect of the coil L is exactly equal to the resistance R. Bulbs $B_1$ and $B_2$ are exactly identical. Answer the following questions based on above information:View Solution
- Which one of the two bulbs lights up earlier, when key K is closed and why?
- What will be the comparative brightness of the two bulbs after sometime if the key K is kept closed and why?