Practice Examination Module 2 Problem 8

Dave Shattuck University of Houston

Note: Units in problem are enclosed in square brackets. Time Allowed: 35 Minutes

Problem Statement: All analog ohmmeters have zero at one end of the scale, and infinity at the other end. Thus, the various scales available are named for their mid-scale reading. The circuit below is intended to be a multi-range ohmmeter. The switch is open for one range, and closed for the other range. The meter indicated is an ammeter. It has a resistance of 50[] and has a full scale current of 1[mA]. Remember that an ohmmeter is calibrated by adjusting RX to give a full-scale reading when the ohmmeter is short-circuited. (Full scale on an ohmmeter is zero.) a) Find the value of the resistor RX. b) Find the mid-scale reading of the ohmmeter, when the switch is open. That is, find the reading when the needle on the scale is in the middle of its range. c) Find the mid-scale reading of the ohmmeter, when the switch is closed. That is, find the reading when the needle on the scale is in the middle of its range. R =500[W] A Switch

Ammeter

RX Terminal 1

+ vS= 2.0[V] -

Terminal 2 Ohmmeter

8.1 Dave Shattuck University of Houston

All analog ohmmeters have zero at one end of the scale, and infinity at the other end. Thus, the various scales available are named for their mid-scale reading. The circuit below is intended to be a multi-range ohmmeter. The switch is open for one range, and closed for the other range. The meter indicated is an ammeter. It has a resistance of 50[] and has a full scale current of 1[mA]. Remember that an ohmmeter is calibrated by adjusting RX to give a full-scale reading when the ohmmeter is short-circuited. (Full scale on an ohmmeter is zero.) a) Find the value of the resistor RX. b) Find the mid-scale reading of the ohmmeter, when the switch is open. That is, find the reading when the needle on the scale is in the middle of its range. c) Find the mid-scale reading of the ohmmeter, when the switch is closed. That is, find the reading when the needle on the scale is in the middle of its range. R =500[W] A Switch

Ammeter

RX Terminal 1

+ vS= 2.0[V] -

Terminal 2 Ohmmeter

a) In the first step, we solve this problem by replacing the ammeter with its equivalent resistance. The key idea after that is that when we short the output, the meter should be full-scale. So, we put a short circuit at the output, and set the current through the ammeter equal to its full-scale reading, 1[mA]. We use this full-scale condition to determine the value of RX.

© Brooks/Cole Publishing Co. Let’s redraw the circuit, replacing the ammeter with its equivalent resistance, and inserting the short circuit. We will call this resistor RM. In addition, let’s label the voltages and currents that would occur at full scale. We have the circuit that follows.

RM= 50[W] RX

1[mA] + vS= 2.0[V] -

Using this circuit, we can solve for RX, using KVL around the loop. This gives us

-vS +1[mA] R M + 1[mA] R X = 0, or

-2.0[V] +1[mA]50[ W ] + 1[mA]RX = 0.

We solve for RX, and we get

1[mA]RX=1.95[V]. Solving for R X , we get

RX =1.95[k W ]. We have solved for this with the switch open. However, even if the switch were closed, this would not change the answer. The circuit with the switch closed, which still requires a short circuit at the terminals of the ohmmeter, would result in the circuit that follows. Notice that the equation that we wrote,

-vS +1[mA] R M +1[m A] R X = 0, is unchanged by the addition of resistor RA.

RA=500[W]

RM= 50[ ] W RX

1[mA] + vS= 2.0[V] -

Thus, our answer remains

RX =1.95[k W ].

b) To solve for the mid-scale value, we need to recognize that the meter is a linear one, so if the needle is in the middle of the range, the current must be half the full-scale value. Thus, we can redraw our ohmmeter circuit with a current of 0.5[mA] indicated. The other key is to recognize that when we do this, we are measuring a resistor value, and we need to insert a resistor at the terminals. Doing so, with the switch open, we have the circuit that follows.

RM= RX= 50[W] 1950[W]

0.5[mA]

+ v = S R 2.0[V] HALF -

We have named the resistance across the terminals RHALF, which is the value we are looking for. Again, we can write KVL, to give

-vS +0.5[mA] R M + 0.5[mA] R X + 0.5[mA] R HALF = 0, or

-2.0[V] +0.5[mA]创 50[ W ]+ 0.5[mA] 1950[ W ] + 0.5[mA]RHALF = 0. Simplifying, we get

0.5[mA ]RHALF=1.0[V]. Solving for R HALF , we get

RHALF =2[k W ]. We note that as has been noted elsewhere, the half scale reading is equal to the resistance of the ohmmeter. We have

RHALF =2[k W ].

Now, for part c), we close the switch, which gives us a new circuit. However, the process is the same; the meter will have half of its full-scale current through it. We have the circuit that follows.

RA=500[W]

RM= 50[ ] W RX=1.95[kW]

0.5[mA]

+ v = S R 2.0[V] HALF2 -

We can solve this circuit for RHALF2, which is the answer we want for this part. To get it, though, we will need to solve for the voltage across it and the current through it, knowing that the ratio of these two quantities is the resistance. Let’s define some variables, in the circuit that follows.

iRA RA=500[W]

+ vRA - RM= 50[W] RX=1.95[kW]

+ 0.5[mA] vX + v = S R 2.0[V] HALF2 iX -

Now, we can write a KVL, and get

-vS +0.5[mA] R M + 0.5[mA]RX +vX = 0, or

-2.0[V] +0.5[mA] 50[ W ]+ 0.5[mA] 1950[ W ] + vX = 0.

Solving for vX, we get

vX =1.0[V]. We can also write KVL to get

-vS +vRA + v X = 0, or

vRA =1.0[V].

From this we can get iRA, by Ohm’s Law,

vRA 1.0[V] iRA = = = 2[mA]. RA 500[W ] Now, using KCL at the upper right hand node, we get

-0.5[mA]- iRA+ i X = 0, or

iX = 2.5[mA].

We now can solve for RHALF2 by Ohm’s Law,

vX 1.0[V] RHALF 2 = = =400[ W ]. iX 2.5[mA]

Please note that the solution given here used more redrawing of the circuit, and more text, than would be expected in an exam solution. This is done for the clarity of the solution. On an exam, it would be expected that you would redraw the circuit about one or two times, so that someone (including you!) could follow your work.