a 15.0k resistor is hooked up to 45.0v battery in a circuit with a switch a.) Draw a circuit diagram for the circuit described.
Label all parts and values. b.) What is the current flowing through the resistor? c.) What is the power dissipated by the resistor?
1 answer:
For item#1 , see attached picture, there is a voltage source, a resistor and wire (where current flow).
For item#2, we use the formula V=IR where "V" is for the voltage, "I" is for the current and "R" for the resistance
Solving for I:
I = v/r
I = 45volts/15k ohms
I=3 mA
For item#3, power is P=VI
p = 45volts * 3ma
P = 0.135 Watts
For item#2, we use the formula V=IR where "V" is for the voltage, "I" is for the current and "R" for the resistance
Solving for I:
I = v/r
I = 45volts/15k ohms
I=3 mA
For item#3, power is P=VI
p = 45volts * 3ma
P = 0.135 Watts
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Answer:
1.dr/dt=0.0096cm/s
2. dA/dt=2.19cm^2/s
Explanation:
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?
for this question, we need to analyze the parameters we know
V=volume of the spherical balloon 1000π cm3
volume of the sphere=
1000π=4/3πr^3
dividing both sides by 4
250*3=r^3
r=9.08cm, the radius of the balloon
dv/dt=dv/dr*dr/dt...................................1
dv/dr ,means
V=
dv/dr=4*pi*r^2
dv/dt=10 cm3/s
from equ 1
10=4*pi*9.08^2*dr/dt
10=1036 dr/dt
dr/dt=10/1036
dr/dt=0.0096cm/s
2. to find the rate at which the area is changing we have,
dA/dt=dA/dr*dr/dt
area of a sphere is 4πr^2
differentiate a with respect to r, radius
dA/dr=8πr
dA/dt=8πr*0.0096
dA/dt=8*pi*9.08*0.0096
dA/dt=2.19cm^2/s
is the rate of change of the surface area