Video Transcript
A galvanometer and a shunt resistor
are connected in parallel to form an ammeter. The resistance of the galvanometer
is 10 times the resistance of the shunt resistor. Which of the lines on the following
graph correctly relates the full-scale deflection current of the galvanometer to the
maximum current measurable by the ammeter? (A) Line I, (B) Line II, (C) Line
III, (D) Line IV, (E) none of the lines.
In this question, we have an
ammeter consisting of a galvanometer connected in parallel with a shunt
resistor. The galvanometer will have a
full-scale deflection current, which we will label as 𝐼 G, and a resistance, which
we will label as 𝑅 G. We are told that the resistance of
the galvanometer is 10 times the resistance of the shunt resistor. So by labeling the resistance of
the shunt resistor as 𝑅 S, we have 𝑅 G equals 10𝑅 S. Let’s clear the answer options for
now to give us some more room.
We can recall that we know an
equation for the required shunt resistance 𝑅 S in order for this ammeter to be able
to measure a maximum current 𝐼. This equation is given by 𝑅 S
equals 𝐼 G 𝑅 G over 𝐼 minus 𝐼 G, where once again 𝑅 S is the shunt resistance,
𝐼 G is the full-scale deflection current on the galvanometer, 𝑅 G is the
galvanometer resistance, and 𝐼 is the maximum current measurable by the
ammeter. By substituting this equation for
the shunt resistance into the relationship 𝑅 G equals 10𝑅 S, we get 𝑅 G equals
10𝑅 S equals 10𝐼 G 𝑅 G over 𝐼 minus 𝐼 G.
We can now rearrange this equation
to see how the maximum current 𝐼 relates to the full-scale deflection current 𝐼
G. We can start by getting rid of the
middle term 𝑅 S here to just leave the resistance 𝑅 G on both sides of the
equation. We can eliminate the resistance by
dividing both sides by 𝑅 G, which gives one equals 10𝐼 G over 𝐼 minus 𝐼 G. We can then multiply both sides by
𝐼 minus 𝐼 G, giving 𝐼 minus 𝐼 G equals 10𝐼 G. Finally, by adding 𝐼 G to both
sides, we get an equation that relates the maximum current 𝐼 with the full-scale
deflection current 𝐼 G. 𝐼 equals 11𝐼 G. So for every one milliamp of
full-scale deflection galvanometer current, we should expect to see 11 milliamps of
maximum ammeter current.
We can now look at the lines on the
graph and see which line correctly relates the full-scale deflection current of the
galvanometer to the maximum current measurable by the ammeter. Lines I, II, and III show that one
milliamp of full-scale deflection current would relate to either one or two
milliamps of maximum measurable current. But we know the relationship should
result in a greater number. Although this graph’s vertical axis
does not go up to 11, we can see that line IV is very steep and thus is the
likeliest to have the relationship we need of one milliamp full-scale deflection of
the galvanometer current giving 11 milliamps of maximum measurable ammeter
current.
To double-check that line IV is
correct, we can divide both sides of this equation by two. Notice, then, that we have one-half
𝐼 equals 11𝐼 G divided by two, or 5.5𝐼 G. Therefore, one-half milliamp of
full-scale deflection current should correspond to 5.5 milliamps of maximum
measurable current. Since the vertical axis of the
graph does go up to 5.5, let’s draw a horizontal line from that value on the axis
over to line IV. Let’s also draw a vertical line
extending from that point on line IV down to the horizontal axis. This helps us see that for line IV,
one-half milliamp of full-scale deflection current does correspond to 5.5 milliamps
of maximum measurable current. And so line IV correctly represents
the relationship between the galvanometer and ammeter current.
Thus, we know that the correct
answer must be option (D). Line IV correctly relates the
full-scale deflection current of the galvanometer to the maximum current measurable
by the ammeter.
