PHY Pre-Lab                Analog Galvanometer          Name:

 

By placing a resistor in parallel or series a galvanometer is creates an ammeter or voltmeter. 

Hint for below…if your voltmeter or ammeter changes the circuit appreciably, your values obtained from the meter will have appreciable error.

 

Ammeter

1.     The resistance added to a galvanometer is

a.     high

b.    low

c.     no extra

 

(Figure 3 in lab manual)

2.    If resistance is added, where?

a.     Series with the part being measured

b.    Parallel with the part being measured

c.     Series with galvanometer

d.    Parallel with galvanometer

 

Voltmeter

1b       The resistance added to a galvanometer is

a.     high

b.    low

c.     no extra

 

(Figure 2 in lab manual)

2b      If resistance is added, where?

  1. Series with the part being measured
  2. Parallel with the part being measured
  3. Series with galvanometer
  4. Parallel with galvanometer

 

3.    When used as an ammeter, if the shunt resistance is ___ than the internal resistance of the galvanometer, the majority of the current will travel through the galvanometer.

a.     lower

b.    higher

c.     unmatched

 

  1. When used as a voltmeter, if the multiplier resistance is too _____ a significant (more than 1/10%) portion of the charge will flow across the galvanometer.
    1. low
    2. high
    3. unmatched

 

5.    What should be the approximate current draw of the shunt resistance?

a.     10 to 15 %

b.    0.1 to 5 %

c.     95 to 99.9 %

d.    85 to 90 %

Figure 3

 

 

6.    What is the voltage drop across “M” compared to the galvanometer?

a.     0.1 to 5 %

b.    95 to 99.9 %

c.     85 to 90 %

d.    10 to 15 %

 

Figure 2

Bonus:  What is the current draw of R1 (% of total)?

 

Below values were obtained from an actual lab; please consult the manual procedures if uncertain how to proceed.

7.    Calculate Ig = V/(10000Ω + r), where V = 5V and r ≈ 63 Ω

 

 

8.    Calculate M = (V / Ig) – r, for a full scale deflection of 3 Volts.

 

  1. Calculate the length of the wire, s = rIg / (I – Ig), if the wire has 0.00053 Ω/cm.