In: Physics
Two light bulbs, one with a resistance of 30 ohms and the other 40 ohms, are arranged in two different circuits with a 120 volt source.
A] The two bulbs are first connected in parallel to the
source.
i. Draw a schematic of the circuit
–clearly labeled.
ii. Determine the current, voltage and power through each
bulb.
B] The two bulbs are now connected in series to the
source.
i. Draw a schematic of the circuit
–clearly labeled.
ii. Determine the current, voltage and power through each
bulb.
C] Rank the bulbs in each situation described, from greatest to least of their brightness.
a. 30ohminparallelcircuit
b. 40 ohm in parallel circuit
c. 30 ohm in series circuit
d. 40 ohm in series circuit
D] Explain how a voltmeter would be connected in each type of connection to verify the voltage calculated for each bulb in part A and B.
E] Explain how an ammeter would be connected to each type of connetion to verify the current calculated for each bulb in part A and B.
Vnet = I * RNet
RNet in parallel = 1/Rnet = 1/30 +1/40
Rnet = 17.14 ohms
Inet = 120/17.14 = 7 Amps
Current Across R1 = V/R1 = 120/30 = 4 AMps
V1 = V2 = 120 Volts
Current across R2 = 120/40 = 3 Amps
Power across R1 (30 ohms in parallel) = I1^2 R1 = 4^2* 30 = 480 Watts
Power across R2 (40 ohms in parallel) = I2^2 R2 = 3^2 * 40 = 360 Watts
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B: when in series
Rnet =R1 + R2
Rnet = 30 + 40 = 70 ohms
in series, current remains same
Inet = I1 = I2 = V/Rnet = 120/70 = 1.714 Amps
V1 = I1R1 = 1.714* 30 = 51.42 VOlts
V2 = I2R2 = 1.714 * 40 = 68.56 Volts
POwer across R1 (30 ohms in series) = P1 = V^1^2/R1 = 51.42^2/30 = 88.1333 Watts
Power across R2 = P2 (40 ohms in series) = V2^2/R2 = 68.56^2/30 = 156.68 Watts
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C.
from above Power values,
we get
a. 30ohminparallelcircuit > b. 40 ohm in parallel circuit >d. 40 ohm in series circuit > c. 30 ohm in series circuit
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D.
Voltmeter mus t be connected in parallel
and
E. AMmeter must be connected in SERIES