In the figure a voltmeter of resistance RV = 439 ? and an ammeter of resistance

Question

In the figure a voltmeter of resistance RV = 439 ? and an ammeter of resistance RA = 2.47 ? 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 = 79.6 ?, what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading, and (c) R'?

Explanation / Answer

resistance of voltmeter, Rv=439 ohms

resistance of ammeter, RA=2.47 ohms

resistance Ro=100 ohms

resistnace R=79.6 ohms

battery emf v=12v

net resistance, Rnet=Ro+(R+RA)*Rv/((R+RA)+Rv)

Rnet=100+(79.6+2.47)*439/(79.6+2.47+439)

Rnet=169.14 ohms

now,

current in battery, i=V/Rnet

i=12/169.14

i=70.95 mA

voltage across Ro is, Vo=100*70.95*10^-3

vo=7.095 v

a)

current through ammeter is,

i'=i*Rv/((R+RA)+Rv)

i'=70.95*10^-3*439/(79.6+2.47+439)

i'=59.8 mA

b)

voltmeter reading v'=v-vo

v'=12-7.095

v'=4.905 v

c)

resistance R'=v'/i'

=4.905/(59.8*10^-3)

=82.02 ohms

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