Download Tutorial 7 - DC Circuits

Tutorial 7 – DC Circuits
26–1 Emf and Terminal Voltage
1. (I) Calculate the terminal voltage for a battery with an internal resistance of 0.900 V and an
emf of 6.00 V when the battery is connected in series with (a) an 81.0-V resistor, and (b) an
810-V resistor.
4. (II) What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4
V when the starter draws 95 A? What is the resistance of the starter? Ans: 0.038 , 0.088 
26–2 Resistors in Series and Parallel
In these Problems neglect the internal resistance of a battery unless the Problem refers to it.
5. (I) A 650-V and a 2200-V resistor are connected in series with a 12-V battery. What is the
voltage across the 2200-V resistor?
16. (II) Determine (a) the equivalent resistance of the circuit shown in Fig. 26–39, and (b) the
voltage across each resistor.
19. (II) What is the net resistance of the circuit connected to the battery in Fig. 26–41?
26–3 Kirchhoff’s Rules
29. (II) For the circuit shown in Fig. 26–47, find the potential difference between points a and b.
Each resistor has R  130 V and each battery is 1.5 V.
33. (II) Determine the magnitudes and directions of the currents through R1 and R2 in Fig. 26–
51.
26–4 Emfs Combined, Battery Charging
42. (II) Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in
Fig. 26–57. If each internal resistance is r  0.450 V , and R  4.00 V , what is the voltage across
the resistor R? Ans: 2.37 V
*55.
(II) A galvanometer has a sensitivity of 45 kV /V and internal resistance 20.0 V . How
could you make this into (a) an ammeter that reads 2.0 A full scale, or (b) a voltmeter reading
1.00 V full scale?
*56.
(II) A galvanometer has an internal resistance of 32 V and deflects full scale for a
55-mA current. Describe how to use this galvanometer to make (a) an ammeter to read
currents up to 25 A, and (b) a voltmeter to give a full scale deflection of 250 V.
Ans: (a) 7.0 × 105  (b) 4.5 × 106 
71. A Wheatstone bridge is a type of “bridge circuit” used to make measurements of resistance.
The unknown resistance to be measured, Rx , is placed in the circuit with accurately known
resistances R1 , R2 , and R3 (Fig. 26–65). One of these, R3 , is a variable resistor which is
adjusted so that when the switch is closed momentarily, the ammeter
shows zero current
flow. (a) Determine Rx in terms of R1 , R2 , and R3 . (b) If a Wheatstone bridge is “balanced”
when R1  630 V ,
resistance?
R2  9 7 2V , and
R3  78.6 V , what is the value of the unknown
85. A potentiometer is a device to precisely measure potential differences or emf, using a “null”
technique. In the simple potentiometer circuit shown in Fig. 26–74, R9 represents the total
resistance of the resistor from A to B (which could be a long uniform “slide” wire), whereas
R represents the resistance of only the part from A to the movable contact at C. When the
unknown emf to be measured, %x , is placed into the circuit as shown, the movable contact C
is moved until the galvanometer G gives a null reading (i.e., zero) when the switch S is
closed. The resistance between A and C for this situation we call Rx . Next, a standard emf,
%s , which is known precisely, is inserted into the circuit in place of %x and again the contact
C is moved until zero current flows through the galvanometer when the switch S is closed.
The resistance between A and C now is called Rs . (a) Show that the unknown emf is given
by
R 
%x   x  %s
 Rs 
where Rx , Rs , and %s are all precisely known. The working battery is assumed to be fresh
and to give a constant voltage. (b) A slide-wire potentiometer is balanced against a 1.0182-V
standard cell when the slide wire is set at 33.6 cm out of a total length of 100.0 cm. For an
unknown source, the setting is 45.8 cm. What is the emf of the unknown? (c) The
galvanometer of a potentiometer has an internal resistance of 35 V and can detect a current
as small as 0.012 mA. What is the minimum uncertainty possible in measuring an unknown
voltage? (d) Explain the advantage of using this “null” method of measuring emf.