In figure 27-62, a voltmeter of resistance R V =310 and an ammeter of resistance

Question

In figure 27-62, a voltmeter of resistance RV =310 and an ammeter of resistance RA =3.23 are being used to measure resistance R in acircuit that also contains a resistance R0 =100 and an ideal battery of emf = 12.0 V. ResistanceR is given by R = V/i, whereV is the voltmeter reading and i is the currentin the resistance R. However, the ammeter reading is noti but rather i', which is i plus thecurrent through the voltmeter. Thus the ratio of the two meterreadings is not R but only an apparent resistanceR' = V/i'. If R = 65.2 ,what are (a) the ammeter reading in milliamperes,(b) the voltmeter reading (in V), and(c) R'?

The figure is the same as in the book. It's for the exact sameproblem you worked out. The values i got were a) 0.0805A b) 4.3369Vc) 53.927

Part C I got right but it is telling me the answers for part a andb are incorrect. Please help me.

Explanation / Answer

   the currents in R and RV are iand (i ' - i) respectively    as V = i R            =(i ' - i) RV    (dividing bothsides with V)    (dividing bothsides with V)    then we get    1 = [(i ' / V) - (i / V)]RV       = [(1 / R ') - (1 / R)]RV    (1 / R) = (1 / R ') - (1 /RV)    R ' = R RV / (R +RV)    the equivalent resistance of thecircuitis    Req = RA +Ro + R '           =RA + Ro + R RV / (R +RV) (a)    the ammeter reading is then    i ' = / Req       = ....... A (b)    the voltmeter reading is    V = - i ' (RA +Ro)          = ........ V (c)    the apparent resistance is    R ' = V/ i '         = ....... (d)    if RV is increased the differencebetween R and R ' decreases    in fact R ' tends to R as RVtends to infinity

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