The circuits, p and q, show two different ammeter-voltmeter methods of measuring resistance. suppose the ammeter has a resistance of 0.10 ohms and the voltmeter has a resistance of 3.0 ohms; the ammeter reads 2.5 amps and the voltmeter reads 6.0 volts. what is the uncorrected value of the resistance in each case? what is the true value of the resistance in each case?

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

R= 12 ohm

2) circuit Q

R = 2.3 ohm

i believe the answer is a hope it and is correct

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form