Name:
Period:
Subject:
Date:
Circuits/Electromagnetics - Worksheet 2
1.
A hot plate has an internal resistance of 22.0 Ù. It operates on 120 V household AC
electricity. (a) How much current did it draw? (b) How much power did it develop? (c) If
it operated for 15 minutes, how much heat did it develop? (d) If a kWh costs 4.5 cents, how
much did it cost to run the thing?
a) I = V/R = 120 V / 22.0 Ù = 5.454545455 A = 5.45 A
b) P = V2 / R = (120 V)2 / 22.0 Ù = 654.5454545 W = 655 W
c) E = P A t = 654.5454545 W A 900 s = 589090.9091 J = 589 000 J or 589 kJ
d) cost = kWh A cost/kWh = 0.6545454545 kW A 0.25 h A 4.5 cents/kWh
= 0.7363636363 cents = 0.736 cents
2.
Examine this useless circuit. What is: (a) the total resistance, ( b) the voltage driving this
circuit, (c) the voltage drop for the 75.0 Ù resistor?
a) RTOT = 100.0 Ù + 50.0 Ù + 75.0 Ù = 225.0 Ù
b) V = RI = 225.0 Ù A 1.20 A = 270 V
c) V = RI = 75.0 Ù A 1.20 A = 90 V
3.
What happens to the light intensity of a set of identical lamps in series when you add an
additional lamp? How come?
The light intensity of a set of identical lamps in series will drop or decrease when you add an
additional lamp in series because the additional lamp is increasing the total resistance. With a
constant voltage source, this means the total current will drop. This will drop the power given
off by the total circuit as well as the power given off by any given lamp (P = I2R)
4.
What happens to the light intensity of a set of identical lamps in parallel when you add an
additional lamp? How come?
The light intensity of a set of identical lamps in parallel will increase when you add an
additional lamp in parallel by the amount of light in the new lamp. The preexisting lamps will
neither dim nor brighten as they are still hooked to the same amount of voltage and drawing
the same current they did before the additional lamp was added. The new lamp will pull
current as well, though, increasing the total load on the power source.
5.
(a) Draw a circuit which has a 120.0 Ù resistor in series with three resistors that are in parallel
with each other; a 25.0 Ù, 35.0 Ù, and 45.0 Ù resistor. The voltage source is a 9.00 V battery.
(b) What is the current for this circuit?
a) See image on right –>
b) RTOT = (25 -1 + 35 -1 + 45 -1) -1 Ù + 120 Ù
= 130.6870229 Ù
I = V/R = 9 V/130.6870229 Ù
= 0.0688668224 A = 0.0688 A or 68.8 mA
6.
What is: (a) the current that goes through the 45.0 Ù resistor and (b) what is the total current
through the circuit?
a) I45 = V45/R = 12.0 V/45.0 Ù
= 0.2666666667 A = 0.267 A or 267 mA
b) RTOT = (25.0 -1 + 15.0 -1 + 45.0 -1 + 35.0 -1) -1 Ù
= 6.350806452 Ù
ITOT = V/RTOT = 12.0 V/6.350806452 Ù
= 1.88952381 A = 1.89 A
7.
What is (a) the voltage drop across the 75 Ù resistor, (b) the total current supplied by the
battery?
R23 = (75 -1 + 55 -1) -1 Ù = 31.73076923 Ù
RTOT = 25 Ù + 35 Ù + 31.73076923 Ù = 91.7 Ù
b) ITOT = V/RTOT = 12.0 V/91.73076923 Ù
= 0.1308176101 A = 0.131 A or 131 mA
a) V75Ù = V23 = R23 I = 4.150943397 V = 4.15 V
8.
What is (a) the total resistance of the circuit, (b) the total current in the circuit, (c) the power
developed in the circuit?
a) RTOT = (112 -1 + 225 -1) -1 Ù + 85.0 Ù
= 159.77744807 Ù = 159.8 Ù
b) V225 = V112 = 112 Ù A 0.250 A = 28.0 V
I225 = V225/R = 28.0 V/225 Ù
= 0.124444444 A = 0.124 A or 124 mA
ITOT = I112 + I225 = 0.250 A + 0.124 A
= 0.374 A or 374 mA
c) P = I2R = (0.374 A)2 A 159.8 Ù = 22.40217901 W = 22.4 W
9.
What is (a) the total resistance of the circuit, (b) the total current in the circuit, (c) the power
developed by R3, the 115 Ù resistor, and (d) the electrical potential provided by the battery?
a) RTOT = (85.0 -1 + 35.0 -1 + 115 -1)-1 Ù + 25.0 Ù = 45.39493294 Ù = 45.4 Ù
b) ITOT = I123 = V123/R123 = 6.00 V/20.39493294 Ù = 0.2941907198 A = 0.294 A or 294 mA
c) P = V2/R = (6.00 V)2/115 Ù = 0.3130434703 W = 0.313 W or 313 mW
d) VBAT = V123 + V4 = V123 + ITOT A R4 = 6.00 V + 0.294 A A 25.0 Ù = 13.354768 V = 13.4 V