Question

In the figure a voltmeter of resistance Rv = 480 s2 and an ammeter of resistance RA = 2.03 12 are being used to measure a resistance R in a circuit that also contains Ro = 100 1 and an ideal battery of emf ε = 12.0 V. Resistance R is given by Vli, 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'li. If R = 89.2 2, what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading, and (c) R'? nin (a) Number Units (b) Number Units Units (c) Number

Answer 1

given , Ro=loon Ru=uson , RA=2.03.1 E=12 V , R=89.2 AA + R A 0 Juction point ALA Apply Kirchoff law. for loop - IA(R) - IA(RA) + IR) = 0 - IA (89.2) – JA (2.03) + Iv (uso). 20 - 91.23 IA + 480 I = o pero 91223 T

for loop 2 & -Iv (Rv) - I (Ro) zo 12 - IN (480) - I (100) 20 (100 I + uso I = 12 pean Applying Juction rule at point "I" In = {bout (I = In tiv h Arqui@ Put the value of I in ean fromegno 100 I tuso lv =12 100 (IATIV) tugo Iv 12 TToo In tS80 IV = 12 ES80I pegn (y)

Selve ogn 0 and egn (4). IA=57.07X103A Iv - 10.84 X10 3A @ Ammeton peading - IA I = 57.07X103A i = 57.07 MA Veltmeter Reiding va v'- 10.84x10 3 x480 203 volt Aus ☺ R'- = 5.2.03 5707X103 IR '= 91.23 n