In the figure a voltmeter of resistance Rv = 442 Ohm and an ammeter of resistanc

Question

In the figure a voltmeter of resistance Rv = 442 Ohm and an ammeter of resistance RA = 3. 17 Ohm are being used to measure a resistance R in a circuit that also contains R0 = 100 fi and an ideal battery of emf = 12. 0 V. Resistance R is given by V/i, where 1/ is the potential across R and i is 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 R' = V/i. If R = 81. 1 Ohm , what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading (in V), and (c) R'? Number Units Number Units Number Units

Explanation / Answer

a) current flowing through ammeter

net resistance in the circuit = Ro + ( Rv parallel (R + Ra))

net resistance = 100 + (442 parallel 84.8)

net resistance = 100 + 71.15 = 171.5

current through the battery = 12/171.5 = 0.07 A

current through ammeter = 0.07 X (Rv/(Rv+R+Ra) ) = 0.0588 A

a) Ammeter reading = 0.0588 A = 58.8 mA

b)vlotemeter reading = current through voltmeter X Rv

current through voltemeter = 0.07 X (R+Ra)/(Rv+R+Ra) = 11.23mA

Voltage reading = 11.23mA X 442 = 4.965 V

c) R = Vmeasured/ Imeasured = 4.965/58.8mA = 84.44 ohms

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