Three resistances of 8O, 16O and 4O are joined in parallel. If each resistor is rated 0.5W , what is the maximum voltage that may be applied to the combination? a.1V b.1.5V c.2V d.2.5V
Answers 2
Answer:
1.5 if three resistance are connect in parallel
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Author:
samcpll
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9
Answer:
Pl look at basic equation
P = VI
I = V/R
Therefore Power = V * (V/R) = V^2/R
V^2 = P/R
Given Power = 0.5 W
R = 98 Ohms
Therefore V^2 = 0.50/98 = 0.0051
Taking Sq root V = sq root(0.0051)
V = 0.0714 Volts
= 71.4 mV
P=I^2*R
2=I^2*120
2/120=I^2
SQRT(2/120)=I
I=0.12909944487 A or 129.1mA
Knowing I and R, you calculate the voltage across it in this case:-
V=I.R
V=0.12909944487*120
V=15.4919333848 V or 15.5V
Those are your exact values for current and voltage with that resistor… to run the rounded numbers:-
P=V*I
P=15.5*0.1291
P=2.00105W
where
initial KE=0
final PE=0
so
initial PE= final KE
initial PE=m*g*h
where
m=mass= 100g which must be in kg (1000g=1kg) so (100g*1kg)/1000g=0.1kg
g= acceleration due to gravity=9.80m/s^2
h= height= unknown
formula for finding the height is
h=L-L*cos theta= L(1-cos theta)
where
L=length of the string=1m
cos theta=cos of 60 degrees=0.5
h=1(1–0.5)
h=0.5m
using the formula
initial PE= final KE
initial PE=m*g*h
initial PE=0.1*9.80*0.5=0.49J
kinetic energy=0.49J
confirming that my answer is correct by using the formula for kinetic energy
K i
Explanation:
answer is b
please mark me as brainliest
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Author:
gill
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Rate an answer:
5