The Ammeter and the Voltmeter Problems and Solutions

 Problem#1

(a) In Fig. 1, what current does the ammeter read if ε = 5.0 V (ideal battery), R₁ = 2.0 Ω, R₂ = 4.0 Ω, and R₃ = 6.0 Ω? (b) The ammeter and battery are now interchanged. Show that the ammeter reading is unchanged.

Fig.1

Answer:
Known:
(a) R₂ and R₃ in combination make

R₂₃ = R₂R₂/(R₂ + R₃) = (4 Ω)(6 Ω)/(10 Ω) = 2.4 Ω

That added to R₁, makes

 R_Total = R₂₃ + R₁ = 4.4 Ω

So total current is

I = ε/R_total = 5 V/4.4 Ω = 1.136 A

This produces a voltage across R₂ and R₃ of

ε₂₃ = IR = (1.136 A)(2.4 Ω) = 2.727 V

That voltage across R₃ produces a current of

I₃ = ε₂₃/R₃ = 2.727 V/6 Ω = 0.45 A

(b) R₁, R₂ in parallel are

R₁₂ = (2 Ω)(4 Ω)/(6 Ω) = 1.33 Ω

in series with R₃ that is

 R_total = R₃ + R₁₂ = 7.33 Ω

So total current is

I = ε/R_total = 5 V/7.33 Ω = 0.68 A

This produces a voltage across R₂ and R₁ of

ε = IR₂₃ = 0.68 A x 1.33 Ω = 0.91 V

That voltage across R₁ produces a current of

I₁ = ε/R₁ = 0.91 V/2 = 0.45 A (Same as before)

Problem#2
In Fig. 2, R₁ = 2.00R, the ammeter resistance is zero, and the battery is ideal. What multiple of  ε/R gives the current in the ammeter?

Fig2

Answer:
The current through the bottom two resistors must be equal. Let us call this current i. The total circuit current is thus 2i. Draw the equivalent circuit which is (2R ∥ R) in series with (R ∥ R). You should find

R_eq = 7R/6

Then Ohm's law with Thevenin's theorem says

(voltage) = (total current)(total equivalent resistance)

For us, this means

ε = (2i)(7R/6)

Therefore

ε/R = 7i/3

Let us now look at the top part of the circuit, and call the current through the 2R branch i₁ and the current through the R branch i₂. Clearly, these currents sum to the total current so

2i = i₁ + i₂

The voltage across the 2R and R resistors are the same, so

i₁(2R) = i₂R

Therefore we can see that i₂/i₁ = 2, which means i₂ makes up 2/3 of the total current, and i₁ makes up 1/3 of the total current. Recall we defined the total current to be 2i so

i₂ = (2/3)(2i) = 4i/3

i₁ = (1/3)(2i) = 2i/3

Finally, convince yourself using the junction rules that the current owing through the ammeter i_A, which is what we want all along, must make up half the difference between i₂ and i₁

i_A= ½ (i₂ ─ i₁)

i_A = ½ [4i/3 ─ 2i/3] = i/3

Since we earlier concluded ε/R = 7i/3 we can assert

ε/7R = i/3 = i_A

Problem#3
In Fig. 3, a voltmeter of resistance R_V = 300 Ω and an ammeter of resistance R_A = 3.00 Ω are being used to measure a resistance R in a circuit that also contains a resistance R₀ = 100 Ω and an ideal battery with an emf of ε = 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. If R 85.0 = Ω, what are (a) the ammeter reading, (b) the voltmeter reading, and (c) R’? (d) If R_A is decreased, does the difference between R’ and R increase, decrease, or remain the same?

Fig.3

Answer:
Here we can analyze the circuit firs to find curr cirrent and voltages across element by relacing the ammeter and voltmeter by their resistance

Here i = i₁ + i₂                                (1)                

KVL 1

ε = Ri₁ + R_Ai₁ + R₀i

12 V = 85.0 Ω x i₁ + 3.00 Ω x i₁ + 100 Ω x i

12 = 88i₁ + 100i

6 = 44i₁ + 50i                       (2)

KVL 2

ε = R_Vi₂ + iR₀

12 V = 300 Ω x i₂ + i x 100 Ω

6 = 150i₂ +  50i                    (3)

From eq. (2) and (3) we get

44i₁ = 150i₂

i₂ = 22i₁/75                           (4)

substitution eq. (1) on (2) and (3) we get

6 = 44i₁ + 50(i₁ + i₂) = 94i₁ + 50i₂ 

3 = 47i₁ + 25i₂;       from (4)

3 = 47i₁ + 25(22i₁/75)

225 = 4075i₁

i₁ = 0.0552 A = 55.2 mA    

then,

i₂ = 22i₁/75 = 0.01619 A = 16.19 mA

and

i = i₁ + i₂ = 71.39 mA

the ammeter reading is i₁ = 0.0552 A = 55.2 mA 
   
(b) the voltmeter reading is V = i₂ x R_v = 16.19 mA x 300 Ω = 4.86 V

(c) R’ = V’/i₁ = 4.86 V/55.2 mA = 87.99 Ω

(d) If R_A is decreased, does the difference between R’ and R is decrease.

Problem#4
When the lights of a car are switched on, an ammeter in series with them reads 10.0 A and a voltmeter connected across them reads 12.0 V (4). When the electric starting motor is turned on, the ammeter reading drops to 8.00 A and the lights dim somewhat. If the internal resistance of the battery is 0.0500 Ω and that of the ammeter is negligible, what are (a) the emf of the battery and (b) the current through the starting motor when the lights are on?

Fig.4

Answer:
V = 12.0 V, when I = 10.0 A with headlights on. When starter motor is engaged, I’ = 8.0 A for the lights. Internal resistance of battery is r = 0.050Ω, with negligible resistance in ammeter. May assume that headlight resistance doesn’t change significantly as current drops from 10 A to 8.0 A.

(a) the emf of the battery

ε – Ir – IR = 0,

where ε – Ir = 12 V.

Therefore,

ε = 12 V + Ir = 12 V + (10 A)(0.050W) = 12.5 V

(b) the current through the starting motor when the lights are on

ε - I’r - (8.0 A)R = 0,

where R = (12 V)/(10 A) = 1.2 W.

I’ is the current from the battery.

I’r = (12.5 V - 9.6 V) = 2.9 V

I’ = (2.9 V)/(0.05 W) = 58 A

The current through the starter motor is I’ – I = 50 A

Problem#5
In Fig. 5, R_s is to be adjusted in value by moving the sliding contact across it until points a and b are brought to the same potential. (One tests for this condition by momentarily connecting a sensitive ammeter between a and b; if these points are at the same potential, the ammeter will not deflect.) Show that when this adjustment is made, the following relation holds: R_x R_sR₂/R₁. An unknown resistance (R_x) can be measured in terms of a standard (Rs) using this device, which is called a Wheatstone bridge.

Fig.5

Answer:
Let i₁ be the current in R₁ and R₂, and take it to be positive if it is toward point a in R₁. Let i₂ be the current in Rs and R_x, and take it to be positive if it is toward b in Rs. The loop rule yields

 (R₁ + R₂)i₁ – (R_x + R_s)i₂ = 0

 Since points a and b are at the same potential,

i₁R₁ = i₂R_s

The second equation gives

i₂ = i₁R₁/Rs

which is substituted into the first equation to obtain

(R₁ + R₂)i₁ = (R_x + R_s)(R₁/R_S)i₁

R_x = R₂R_s/R₁   

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