In the figure at the right, a voltmeter of resistance R_v = 350 Ohm and an ammeter of resistance R_A = 2.50 Ohm are being used to measure a resistance R = 85.0 Ohm in a circuit that also contains a resitance R_0 = 75.0 Ohm and an ideal battery of emf epsilon = 120 V. Resistance R is given by R = 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. Determine the voltmeter reading Determine the ammeter reading Determine R\'.
Total circuit resistance = 75+(85+2.50)//(350)
=75+87.50*350/(87.50+350)=145 ohm
Current from battery = 12/145=0.0827 A
Volt Drop across Ro=75*0.0827=6.20 V
Voltmeter reading is thus 12-6.20=5.80 V..........................(a)
Current share thru\' ammeter branch=
0.0827*350/[87.50+350)]=0.06616 A..........................(b)
R\'=5.80/0.06616=87.66 ohm.................................(c)
Homework Sourse
