In the figure a voltmeter of resistance = 410 omega and an ammeter of resistance

Question

In the figure a voltmeter of resistance = 410 omega and an ammeter of resistance R_A = 1.92 omega are being used to measure a resistance R in a circuit that also contains R0 = 100 omega and an ideal battery of emf epsilon = 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 = 118 omega, what are (a) the ammeter reading in milliamperes, (b) the voltmeter reading, and (c) R'?

Explanation / Answer

Find the equivalant resistance of R, RA and RV
Let this be R'

R' = (R+RA)||RV
= (118 + 1.92) ||410
=119.92 || 410
= 119.92*410 / (119.92+410)
=92.8 ohm

this R' and Ro are in series

I net = E/(R'+Ro)
I net = 12 / (92.8 + 100)
i net = 0.062 A

Voltmeter reading = V across R'
= inet * R'
= 0.062 * 92.8
= 5.78 V

Ammeter reading = V / (R+RA)
= 5.78 / (118 + 1.92)
= 0.048 A
Answer:
a) 0.048 A
b) 5.78 V

what is R'???

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