In the figure a voltmeter of resistance R V = 456 ? and an ammeter of resistance
ID: 2245050 • Letter: I
Question
In the figure a voltmeter of resistance RV = 456 ? and an ammeter of resistance RA = 1.44 ? are being used to measure a resistance R in a circuit that also contains R0 = 100 ? and an ideal battery of emf ? = 12.0 V. Resistance R is given by V/i, where V 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 = 92.5 ?, what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading (in V), and (c) R'?
Explanation / Answer
a) total equivalent resistance of circuit = 1.44 + 100 + ( (92.5+1.44) * 456 ) / ( (92.5+1.44) + 456 ) = 102.44 ohm...
so.. current through circuit = 12 / 102.44 = 0.11714174 A
so.. current through ammeter = 0.11714174 * 456 / ( (92.5+1.44) + 456 ) = 0.09713174914953 A = 97.13174915 mA
so.. ammeter reading = 0.09713174915 mA
b) current throuch voltmeter = 0.11714174 - 0.09713174914953 = 0.02000999085047 A
so... voltmeter reading = 456 * 0.02000999085047 = 9.12455582781432 V
c) R' = 9.12455582781432 / 0.09713174914953 = 93.93999292412 V
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