Download Recitation #9 Solution

Physics 212 Spring 2000
Recitation Activity #9: DC Circuits
NAME: Solution
Instructor: Redwing
DATE: March 9, 2000
This activity is based on the following concepts:
 Loop rule: follow a closed loop around any circuit; the net sum of the changes in potential
that you encounter = 0;
 Junction rule: the algebraic sum of currents at a junction in a circuit = 0;
 Resistances in series: net resistance = sum of all resistors;
 Resistances in parallel: net reciprocal resistance = sum of all reciprocal resistances
 The resistance of an ideal ammeter is zero, and it is essential in real ammeters that the
resistance of the instrument be very small compared to other resistances in the circuit.
 The resistance of an ideal voltmeter is infinity, and it is essential in real voltmeters that the
resistance of the instrument be very large compared to the resistance of any circuit elements
across which it is connected.
i2
Exercise 1. Consider the circuit shown
below. Use the junction and loop rules to
determine the currents through the 3 resistors.
In your solution, remember to specify the
DIRECTION of each current as well as its
magnitude.
10 
9V
20 
i1
i3
9V
9V
20 
i1 – i2 – i3 = 0

i1 = i2 + i3
9 V – (10 ) i2 – 9 V + (20 ) i3 = 0
9 V – (10 ) i2 – (20 ) i1+ 9 V = 0
18 V – (10 ) i2 – (20 ) 1.5 i2= 0
i2= 0.45 A

i1 = 0.675 A 


i2 = 2 i3
i1 = 1.5 i2
i3 = 0.225 A
Exercise 2. An instrument used to measure current is called an
ammeter. The ammeter is connected into a circuit in series in
the loop in which the desired current reading is required. The
resistance of an ideal ammeter is zero, and it is essential in real
ammeters that the resistance of the instrument be very small
compared to other resistances in the circuit. Why?
Since the current is i = V/RTotal, if the resistance of the
ammeter changes RTotal then the current measurement will
be effected. Since RA is in series, if RA is large compared to
the rest of the resistances in the circuit then it will impact
RTotal.
+
-

R2
r
A
R1
V
A meter to measure potential difference is called a voltmeter. To find the potential difference
between two points in a circuit the voltmeter terminals are connected between those two
points. The meter is connected in parallel with the circuit elements between the two points.
The resistance of an ideal voltmeter is infinity, and it is essential in real voltmeters that the
resistance of the instrument be very large compared to the resistance of any circuit elements
across which it is connected. Why?
If the resistance of a voltmeter is not large compared to R1, then current will
shunt through the voltmeter and the voltage drop across R1, V = iR1, will be
affected.
What would be the effect of a voltmeter that was really an ideal ammeter mistakenly
connected in parallel?
Due to the low resistance of the ammeter, the current will shunt by R1 and
drastically change the voltage drops in the circuit.
In the circuit for exercise 2, assume that  = 5.0 V, r = 2.0 , R1 = 5.0  and R2 = 4.0 . If
the ammeter resistance is RA = 0.1  what percentage error is made in reading the current?
Assume the voltmeter is not present for this problem.

5.0V
 0.4505 A
r  R1  R2  R A (2  5  4  0.1)

5.0V
i

 0.4545 A
r  R1  R2 (2  5  4)
i 0.004

 0.0089 
0.9 %
i
0.4545
iA 

In the circuit for exercise 2, assume that  = 3.0 V, r = 100 , RA = 0.1 , R1 = 250  and
R2 = 300 . If the voltmeter resistance RV = 5 k, what percentage error is made in reading
the potential difference across R1 if the voltmeter reads 1.12 V?

3.0V
i

 4.61mA
r  R1  R2  R A (100  250  300  0.1)
V1 = iR1 = 1.15 V
V 1.15V  1.12V

 0.026
2.6 %
V1
1.15V
Scoring Guidelines:
100
90
75
60
40
0
All correct
miss one
some correct
none to few correct, but shows effort
Turned in paper, but no real effort
No paper