A simple ohmmeter is made by connecting a 4.5 V battery in series with a resista

Question

A simple ohmmeter is made by connecting a 4.5 V battery in series with a resistance R and an ammeter that reads from 0 to 1.00 mA, as shown in the figure below. Resistance R is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of 1.00 mA.

(a) What external resistance across the leads results in a deflection of 10% of full scale?

(b) What resistance results in a deflection of 50% of full scale?

(c) What resistance results in a deflection of 90% of full scale?

A simple ohmmeter is made by connecting a 4.5 V battery in series with a resistance R and an ammeter that reads from 0 to 1.00 mA, as shown in the figure below. Resistance R is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of 1.00 mA. What external resistance across the leads results in a deflection of 10% of full scale? What resistance results in a deflection of 50% of full scale? What resistance results in a deflection of 90% of full scale? If the ammeter has a resistance of 34.0 and the internal resistance of the battery is negligible, what is the value of R?

Explanation / Answer

I1 = E/R = 1mA

R = E/I1 = 4.5/0.001 = 4.5*10^3 = 4500 ohms

a) I2 = 0.1*I1 = 0.1*10^-3
I2 = E/(R+R2)

R1 = 40500 ohms

b) I3 = 0.5mA

I3 = E/(R+R3)

R3 = 4500 ohms

c) I4 = 0.9mA


I4 = E/(R+R4)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.