771_PartUniversity Physics Solution

# 771_PartUniversity Physics Solution - Direct-Current...

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Direct-Current Circuits 26-13 (b) For a 500-mV voltmeter, the resistances are in series and the current is the same through each: ( ) cs ab VI R R = + and 3 sc 6 500 10 V 25.0 975 . 500 10 A ab V RR I × =− = Ω × EVALUATE: The equivalent resistance of the voltmeter is eq s c 1000 R = += Ω . The equivalent resistance of the ammeter is given by eq sh c 111 R =+ and eq 0.625 R = Ω . The voltmeter is a high-resistance device and the ammeter is a low-resistance device. 26.32. IDENTIFY: The galvanometer is represented in the circuit as a resistance R c . Use the junction rule to relate the current through the galvanometer and the current through the shunt resistor. The voltage drop across each parallel path is the same; use this to write an equation for the resistance R . SET UP: The circuit is sketched in Figure 26.32. Figure 26.32 We want that a 20.0 A I = in the external circuit to produce fs 0.0224 A I = through the galvanometer coil. EXECUTE: Applying the junction rule to point a gives af ss h 0 III −= sh a fs 20.0 A 0.0224 A 19.98 A II I =−= = The potential difference ab V between points a and b must be the same for both paths between these two points: () fs c sh sh IRR IR ( ) sh sh c fs 19.98 A 0.0250 9.36 22.30 9.36 12.9 0.0224 A IR I Ω = Ω Ω = Ω EVALUATE: sh c ; R << + most of the current goes through the shunt. Adding R decreases the fraction of the current that goes through c . R 26.33. IDENTIFY: The meter introduces resistance into the circuit, which affects the current through the 5.00-k resistor and hence the potential drop across it. SET UP: Use Ohm’s law to find the current through the 5.00-k resistor and then the potential drop across it. EXECUTE: (a) The parallel resistance with the voltmeter is 3.33 k , so the total equivalent resistance across the battery is 9.33 k , giving I = (50.0 V)/(9.33 k ) = 5.36 mA. Ohm’s law gives the potential drop across the 5.00-k resistor: V 5 k = (3.33 k )(5.36 mA) = 17.9 V (b) The current in the circuit is now I = (50.0 V)/(11.0 k ) = 4.55 mA. V 5 k = (5.00 k )(4.55 mA) = 22.7 V. (c) % error = (22.7 V – 17.9 V)/(22.7 V) = 0.214 = 21.4%. (We carried extra decimal places for accuracy since we had to subtract our answers.) EVALUATE: The presence of the meter made a very large percent error in the reading of the “true” potential across the resistor. 26.34. IDENTIFY: The resistance of the galvanometer can alter the resistance in a circuit. SET UP: The shunt is in parallel with the galvanometer, so we find the parallel resistance of the ammeter. Then use Ohm’s law to find the current in the circuit. EXECUTE: (a) The resistance of the ammeter is given by 1/ R A = 1/(1.00 ) + 1/(25.0 ), so R A = 0.962 . The current through the ammeter, and hence the current it measures, is I = V / R = (25.0 V)/(15.96 ) = 1.57 A. (b) Now there is no meter in the circuit, so the total resistance is only 15.0 . I = (25.0 V)/(15.0 ) = 1.67 A (c) (1.67 A – 1.57 A)/(1.67 A) = 0.060 = 6.0% EVALUATE: A 1- shunt can introduce noticeable error in the measurement of an ammeter.

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771_PartUniversity Physics Solution - Direct-Current...

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