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R
R
R
R
Circuit A Circuit B
I = 3 A
CIRCUITS WORKSHEET
- Determine the equivalent (total) resistance for each of the following circuits below.
- Determine the total voltage (electric potential) for each of the following circuits below.
13V 12 V
- In a series circuit there is just one path so the charge flow is constant everywhere (charge is not lost or gained). Circuit B was made by adding 2 more identical resistors in series to circuit A
a) How is the charge flow out of the battery (and back into it) affected by adding more bulbs in
series? Charge flow or current decreases as
total resistance increases
b) If the resistors were light bulbs, how do you expect the brightness of the bulbs to be affected by adding more bulbs in series?
Brightness gets dimmer since less current or charge passing through each bulb AND
smaller voltage drop across each bulb (the voltage gain at the battery is now distributed
among 3 bulbs as opposed to just one).
c) How is the brightness in the 2 circuits related to charge flow or current?
The brightness is directly related to current since the less charge flowing through each
bulb each second, the less energy/charge is lost and converted to light
d) How does the current in circuit B compare to circuit A? Circuit B has three times the total resistance (same V) so current supplied by battery drops three fold. Circuit B would have only 1A of current.
1 2 3
eq
eq
R
R R R R
1 2
eq
eq
R
R R R
1 2 3
eq
eq
R
R R R R
R 1
R 2
R 3
6 A 2A
2A
2A
2A
e) How is current (I) related to the resistance of the circuit?
The current is inversely related to the total resistance of the circuit (Ohm’s Law)
f) If the resistance of a circuit is quadrupled, by what factor does the current change? 1/4th
g) Fill out the table for the circuit diagramed at the right.
h) Is there a relationship between resistance and voltage drop in a series circuit? If so, state it.
Ohm’s Law: V = IR
c) If the resistors were light bulbs, explain in terms of charge flow (current) and energy per charge (voltage) which bulb would be brightest / dimmest. The brightness of the bulb is related to
- V (amount of electrical energy lost and converted to heat/light) and
- I (the higher the current through the bulb, the more charge per sec converting energy to light)
In this case, the current through each resistor is the same, so V determines the brightness. Since the voltage drop across the 30 resistor is greatest, it would be the brightest bulb. Conversely, since the voltage drop across the 10 resistor is least, it would be the dimmest bulb.
- In a parallel circuit, there is more than one loop or pathway so charge flow gets split up or recombined at junction points. Therefore current is not the same at every point in the circuit
a) How does the current through the one resistor in circuit A, compare to the current through each resistor in circuit B? (Use Kirchoff Loop rule on circuit B to look at the current in each path.) The current through each resistor in circuit B is the same as the current through the resistor in circuit A (I = V/R. V across R in circuit B is same as circuit A) b) How does the sum of the currents through the three bulbs in circuit B compare to current from the battery in circuit A? Since the current across each bulb in circuit B is the same as in circuit A and there are three pathways, the sum of the currents in B is 3x current in circuit A
Circuit Position
Voltage (V)
Current (A)
Resistance (Ω) Power (W)
Total 6.00 0.10 60 0.
I = ____
6V
Circuit B
Circuit A
6V
I = ____
I
I 1 I 2 I 3 I 4
I Req
12 V
out. Since R2 and R3 are identical, the current splits equally down the R2 and R3 pathway (I 2 = I 3 ). At the next branch point, the current recombines to the original total current and this is what goes through R1.
I 2 = I 3 < I 1
b) Imagine that the resistors in parallel (R 2 and R 3 ) were a single resistor. How would the combined equivalent resistance of R 23 compare to the resistance of R 1? The resistors in the box are in parallel so the equivalent resistance is
The equivalent resistance is half of R
c) Now let R 1 = 10 , R 2 = 20 and R 3 = 30 Fill out the table for the circuit
Questions 6 and 7 refer to the following: The diagram to the right represents an electric circuit consisting of four resistors and a 12-volt battery
What is the equivalent resistance of the circuit shown?
What is the current measured by ammeter A shown in the diagram? KLR on loop with 12V battery and 6 resistor KLR
Circuit Position
Voltage (V) Current (A)^
Resistance (Ω)
Power (W)
Total 6.
(current thru battery)
22 (Req)
1 112 R R / 2
R (^) eq R R R eq
1 2 3 4
eq
eq
R
R R R R R
I A
I
V
V V
V
bat
1
1
6
6
- A 6.0-ohm lamp requires 0.25 ampere of current to operate. In which circuit below would the lamp operate correctly when switch S is closed?
It would only operate in C. In A and D, closing the switch would introduce a pathway of zero resistance. ALL of the current would go down the path of no resistance leaving NO current passing through the lamp (the circuit would be short circuited). In B, once the switch was closed and the circuit included the battery, ALL of the current would go down the path of no resistance leaving no current passing through the lamp (short circuit).
Questions 9 and 10 refer to the following: A 50.-ohm resistor, an unknown resistor R , a 120-volt source, and an ammeter are connected in a complete circuit. The ammeter reads 0.50 ampere.
Calculate the equivalent resistance of the circuit shown.
Determine the resistance of resistor R shown in the diagram. Resistors in series OR could use KLR
50
50
R
R
I IR
V V
V V V
V
R
bat R
I
eq
eq
bat eq
R
R
V IR
1 2
R
R
Req R R
1 2
eq
eq
R
R R R
R 2 R 3 R 4
R 1
R 234
R 1
Req = 1333
P=800 W
I=P/V = 800/110 = 7.27A
Microwave
Toaster
R 1
R 2
R 3
Coffee Maker
V=110 V
P=100 W
I=P/V = 1000/110 = 9.09A
P=1800 W
I=P/V = 1800/110 = 16.4A
20A Fuse
When the microwave and coffeemaker are on, a total current of 23.7A=7.27+16.4A is drawn from power supply. This exceeds the max current of the 20A fuse so it will melt and the current through the circuit will stop.
- Determine the equivalent resistance of two resistors of 12 and 18 when they are connected
a) in parallel b) in series
- Assume that you have five one thousand ohm (1000 ) resistors. In series, 1000 would make Req in multiples of 1000 In parallel, 1000 would make 1000/2 (for 2 in parallel), 1000/3 (for 3 in parallel), 1000/4 (for 4 in parallel), … a) Devise a circuit with an equivalent resistance of 1333 ohms. Need 3 in parallel (Req=333 W) which is in series with another
1 2
eq
eq
R
R R R
eq
eq R
R
R 12 = 500
R 3456 = 250
b) Devise a circuit with an equivalent resistance of 750 ohm. Cant do unless have six 1000 resistors (500 with 2 in parallel) in series with (250 made with 4 in parallel)
c) Using all five resistors, what is the smallest resistance that can be constructed? Smallest Req would be with all 5 in parallel
d) Using all five resistors, what is the largest resistance that can be constructed? Largest Req would be with all 5 in series
- Given the circuit at right:
a) Draw the current flow on the circuit. Label each current
Req 1000 1000 1000 1000 1000 5000
3
15
6
24 3
10 (^) 8
5 I 1 =3A
I 2
I 3
I 4 I 5
I 6 I 7
I 2
I 3
I 1
c) Complete the table
R ( ) V (v) I (A)^ P (W)
Battery 39.
3 9.0 3 27
