In Fig. 27-58, a voltmeter of resistance RV = 443 ? and an ammeter of resistance
ID: 2154874 • Letter: I
Question
In Fig. 27-58, a voltmeter of resistance RV = 443 ? and an ammeter of resistance RA = 3.37 ? 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 = 116 ?, what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading (in V), and (c) R'?
Explanation / Answer
sol take help Total circuit resistance = 100+(103+2.51)//(258) =100+105.51*258/(105.51+258)=174.885 ohm so Current from battery = 12/174.885=0.0686 A Volt Drop across Ro=100*0.0686=6.86 V then Voltmeter reading is thus 12-6.86=5.14 V..........................(b) so Current share thru' ammeter branch= 0.0686*258/[105.51+258)]=0.0487 A..........................(a) then R'=5.14/0.0487=105.544 ohm.................................(c)