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Chapter 18: Problem 77

A galvanometer has a coil resistance of \(34.0 \Omega .\) It is to be made into a voltmeter with a full-scale deflection equal to \(100.0 \mathrm{V}\). If the galvanometer deflects full scale for a current of \(0.120 \mathrm{mA},\) what size resistor should be placed in series with the galvanometer?

Short Answer

Expert verified
Answer: The size of the resistor should be approximately 833299.33 Ω.

Step by step solution

01

Determine the current needed for full-scale deflection

The current required to deflect the galvanometer full scale is given as \(0.120 \mathrm{mA}\). Let's convert it to Amperes to be consistent with the unit: \(0.120 \mathrm{mA} = 0.000120 \mathrm{A}\).
02

Calculate the total resistance of the voltmeter

Using Ohm's Law, we can calculate the total resistance needed for the voltmeter to have a full-scale deflection equal to \(100.0 \mathrm{V}\): \(V = IR\) \(R_{total} = \frac{V}{I}\) Plug in the values: \(R_{total} = \frac{100.0 \mathrm{V}}{0.000120 \mathrm{A}}\) \(R_{total} = 833333.33 \Omega\)
03

Calculate the size of the resistor to be placed in series with the galvanometer

Now, we have the total resistance needed, and we know the coil resistance of the galvanometer. The resistor required in series should bring the total resistance to the required value, so we just subtract the galvanometer's coil resistance from the total resistance: \(R_{resistor} = R_{total} - R_{coil}\) \(R_{resistor} = 833333.33 \Omega - 34.0 \Omega\) \(R_{resistor} = 833299.33 \Omega\) The size of the resistor to be placed in series with the galvanometer should be approximately \(833299.33 \Omega\).

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Most popular questions from this chapter

An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
(a) What are the ratios of the resistances of (a) silver and (b) aluminum wire to the resistance of copper wire $\left(R_{\mathrm{Ag}} / R_{\mathrm{Cu}} \text { and } R_{\mathrm{A}} / R_{\mathrm{Ca}}\right)$ for wires of the same length and the same diameter? (c) Which material is the best conductor, for wires of equal length and diameter?
A \(12-\Omega\) resistor has a potential difference of \(16 \mathrm{V}\) across it. What current flows through the resistor?
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We can model some of the electrical properties of an unmyelinated axon as an electric cable covered with defective insulation so that current leaks out of the axon to the surrounding fluid. We assume the axon consists of a cylindrical membrane filled with conducting fluid. A current of ions can travel along the axon in this fluid and can also leak out through the membrane. The inner radius of the cylinder is \(5.0 \mu \mathrm{m} ;\) the membrane thickness is \(8.0 \mathrm{nm} .\) (a) If the resistivity of the axon fluid is \(2.0 \Omega \cdot \mathrm{m},\) calculate the resistance of a 1.0 -cm length of axon to current flow along its length. (b) If the resistivity of the porous membrane is \(2.5 \times 10^{7} \Omega\).m, calculate the resistance of the wall of a 1.0 -cm length of axon to current flow across the membrane. (c) Find the length of axon for which the two resistances are equal. This length is a rough measure of the distance a signal can travel without amplification.
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