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Chapter 27: 10Q (page 794)

Cap-monster maze.In Fig. 27-22, all the capacitors have a capacitance of60μC, and all the batteries have an emf of 10 V. What is the charge on capacitor C? (If you can find the proper loop through this maze, you can answer the question with a few seconds of mental calculation).

Short Answer

Expert verified

The charge on capacitor C is60μC.

Step by step solution

01

Step 1: Given

V=10VC=6.0μF

02

Determining the concept

Here, use the formula for the charge in terms of the voltage and the capacitor. When the capacitors are connected in series with battery, voltage remains same through the capacitors.

Formulae are as follow:

q=CV

Where, C is capacitance,V is potential difference, q is charge

03

Determining the charge on capacitor C

Consider the loop as shown in figure 27-22 in which the capacitor C is connected in series with the battery shown by the path in red.

So, the voltage across capacitor C is 10 V

Now, the charge is as follow,

q=CVq=6.0×10q=60μC

Hence, the charge on capacitor C is60μC

Therefore, first find the voltage through the capacitors. From that, find the charge through the capacitors.

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Most popular questions from this chapter

In Fig. 27-19, a circuit consists of a battery and two uniform resistors, and the section lying along an xaxis is divided into five segments of equal lengths.

(a) Assume thatR1=R2and rank the segments according to the magnitude of the average electric field in them, greatest first.

(b) Now assume thatR1>R2and then again rank the segments.

(c) What is the direction of the electric field along the xaxis?

The figure shows a section of a circuit. The resistances are R1=2.0Ω , R2=4.0Ωand R3=6.0Ω, and the indicated current is I=6.0A . The electric potential difference between points A and B that connect the section to the rest of the circuit is VA−VB=78V . (a) Is the device represented by “Box” absorbing or providing energy to the circuit, and (b) At what rate?

A 120V power line is protected by a 15A fuse. What is the maximum number of 500W lamps that can be simultaneously operated in parallel on this line without “blowing” the fuse because of an excess of current?

In Fig. 27-58, a voltmeter of resistance RV=300Ωand an ammeter of resistanceRA=3.00Ω are being used to measure a resistance R in a circuit that also contains a resistance R0=100Ωand an ideal battery with an emf of ε=12.0V. ResistanceR is given by R=V/i, whereV is the potential across Rand iis the ammeter reading. The voltmeter reading is V', which is V plus the potential difference across the ammeter. Thus, the ratio of the two-meter readings is not R but only an apparent resistance role="math" localid="1664348614854" R'=V/i. If R=85.0Ω, what are (a) the ammeter reading, (b) the voltmeter reading, and (c)R' ? (d) IfRA is decreased, does the difference betweenR' andR increase, decrease, or remain the same?

You are to connect resistors R1and R2, withR1>R2, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.
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