In the figure at the right, a voltmeter of resistance R V = 350 ? and an ammeter
ID: 1446655 • Letter: I
Question
In the figure at the right, a voltmeter of resistance RV = 350 ? and an ammeter of resistance RA = 2.50 ? are being used to measure a resistance R = 85.0 ? in a circuit that also contains a resistance R0 = 75.0 ? and an ideal battery of emf E = 12.0 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.
(a) Determine the voltmeter reading.
(b) Determine the ammeter reading.
(c) Determine R'.
These are my teachers requirements if you could please follow them I would really appreciate it thank you :)
In addition to being neat and clear, and actually answering the question, you must:
1) show the original principle (often in equation form)
2) substitute variables as needed?
3) solve it first (before substituting numbers)
4) show every substitution (with units and correct sig figs)
5) present a boxed answer (with units and correct sig figs)
In a Force problem in more than one dimension, you MUST start with a Free Body Diagram.
Explanation / Answer
Rv = 350 ohm
Ra = 2.5 ohm
R = 85 ohm
Ro = 75 ohm
E = 12V
a) let total current through the battery be I, current through voltmeter be i
Net resistance across EMF source be r
Now, R, Ra are in series so net resistance = R + Ra = 87.5 ohm
This is parallel to Rv, so net resistance = (1/87.5 + 1/350)^-1 = 70 ohm
This is in series with Ro, so r = Ro + 70 = 145 ohm
Hence, I = E/r = 0.0827 A
V across voltmeter = E -IRo = 12- 0.0827*75 = 5.793 V
b) Voltage across Ra and R combined = 5.793V
Ia = 5.793/[Ra+R] = 0.0662 A
c) R' = V'/i = 5.793/0.0662 = 87.50755 ohms