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How is the initial current through a bulb affected by putting a capacitor in series in the circuit? Explain briefly.

Short Answer

Expert verified

In the presence of a capacitor, the direct current, flowing in the circuit, starts reducing and becomes zero after a certain time interval.

Step by step solution

01

Parallel Plate Capacitor

When two conducting plates are placed parallel to each other, with a layer of insulating material between them, the combination is called a parallel plate capacitor. It has the ability to store energy in the form of an electric field, between the plates.

02

Explanation

When a capacitor is connected to a d.c. source, the charge starts accumulating at the plates. Due to this accumulation of charge, an electric field is developed between the plates, such that it opposes the flow of current through the circuit.

With more and more charge accumulation, the strength of the electric field between the plates also increases. Thus, it will more strongly oppose the current. So, the current in the circuit will be reduced.

03

Conclusion

When a capacitor is connected to a d.c. source, the current in the circuit starts reducing and becomes zero after a certain time interval.

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Most popular questions from this chapter

Question: in circuit 1 (Figure 19.72), an uncharged capacitor is connected in series with two batteries and one light bulb. Circuit 2 (Figure 19.72) contains two light bulbs identical to the bulb in the circuit; in all other respects, it is identical to circuit 1. In circuit 1, the light bulb stays lit for 25 s. The following questions refer to these circuits. You should draw diagrams representing the fields and charges in each circuit at the times mentioned, in order to answer the questions.

(a)One microsecond after connecting both circuits, which of the following are true? Chose all that apply: (1) the net electric field at location B in circuit 1 is larger than the net electric field at location B in circuit 2. (2) At location A in 1, electrons flow to the left. (3) At location A in circuit 1, the electric fields due to charges on the surface of the wires and batteries points to the right. (4) in circuit 1 the potential difference across the capacitor plates is equal to the emf of the batteries. (5) The current in circuit 1 is larger than the current in circuit 2.

(b)Two seconds after connecting both circuits, which of the following are true? Choose all that apply: (1) there is more charge on the plates of capacitor 1 than there is on the plates of capacitor 2. (2) there is negative charge on the right plate of the capacitor in circuit 1. (3) At location B in circuit 2 the net electric field points to the right. (4) At location B in circuit 2 the fringe field of the capacitor points to the right. (5) At location A in circuit 1 the fringe field of the capacitor points to the left.

(c)Which of the graphs in Figure 19.73 represents the amount of charge on the positive plate of the capacitor in circuit 1 as a function of time?

(d)Which of the graphs in Figure 19.73 represents the current in circuit 1 as a function of time?

State whether the following statement is true or false, and briefly explain why: โ€œIn the two circuits shown in Figure 19.64, the battery output power is greater in circuit 2 because there is an additional resistor dissipating power.โ€

Two resistor each with resistance of 4ร—106ฮฉ are connected in series to a 60 V power supply whose internal resistance is negligible. You connect the voltmeter across one of these resistors and this voltmeter has an internal resistance of 1ร—106ฮฉ. What is the reading on the voltmeter?

Using thick connecting wires that are very good conductors, a Nichrome wire (โ€œwire 1โ€) of length L1 and cross-sectional area A1 is connected in series with a battery and an ammeter (this is circuit 1). The reading on the ammeter is I1. Now the Nichrome wire is removed and replaced with a different wire (โ€œwire 2โ€), which is 2.5 times as long and has 5.5 times the cross-sectional area of the original wire (this is circuit 2). In the following question, a subscript 1 refers to circuit 1, and a subscript 2 refers to circuit 2. It will be helpful to write out your solutions to the following questions algebraically before doing numerical calculations. (Hint: Think about what is the same in these two circuits.)(a) What is the value of I2/ I1, the ratio of the conventional currents in the two circuits? (b) What is the value of R2/ R1, the ratio of the resistances of the wires? (c) What is the value of E2/ E1, the ratio of the electric fields inside the wires in the steady states?

A long Iron slab of width w and height h emerges from a furnace, as shown in Figure 19.79. Because the end of the slab near the furnace is hot and the other end Is cold, the electron mobility increases significantly with the distance x. The electron mobility is u=u0+kxwhereu0is the mobility of the iron at the hot end of the slab. There are n iron atoms per cubic meter, and each atom contributes one electron to the sea of the mobile electron (we can neglect the small thermal expansion of the iron). A steady state conventional current runs through the slab from the hot end towards cold end, and an ammeter (not shown) measures the current to have a magnitude I in amperes. A voltmeter is connected to two locations a distance d apart, as shown. (a) Show the electric field inside the slab at two locations marked withร—. Pay attention to the relative magnitudes of the two vectors that you draw. (b) Explain why the magnitude of the electric field is different at these two locations. (c) At a distance x from the left voltmeter connection, what is the magnitude of the electric field in terms x and the given quantities w,h,d,u0,k,l, and n ( and fundamental constants)? (d) What is the sign of potential difference displayed on the voltmeter? Explain briefly. (e) In terms of the given quantities w,h,d,u0,k,l, and n and ( and fundamental constants), what is the magnitude of the voltmeter reading? Check your work. (f) What is the resistance of this length of the iron slab?

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