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24 Jun 2015, 00:16
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The amount of electrical current that flows through a wire is inversely proportional to the resistance in that wire. If a wire currently carries 4 amperes of electrical current, but the resistance is then cut to one-third of its original value, how many of electrical current will flow through the wire?
A. 3/4 amperes
B. 4/3 amperes
C. 4 amperes
D. 8 amperes
E. 12 amperes
A. 3/4 amperes
B. 4/3 amperes
C. 4 amperes
D. 8 amperes
E. 12 amperes
24 Jun 2015, 02:06
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Let I1 be the initial current and R1 be the initial resistance.
I1 is inversely proportional to R1
i.e I1 = k/ R1 ; where k is some constant.
From the given data:
4 = k / R1
k = 4R1
Given R2 = R1/3
I2
= k/R2
= k/ R1/3
= 3k/ R1
= 3 * 4* R1/ R1
= 12 A
Answer E
I1 is inversely proportional to R1
i.e I1 = k/ R1 ; where k is some constant.
From the given data:
4 = k / R1
k = 4R1
Given R2 = R1/3
I2
= k/R2
= k/ R1/3
= 3k/ R1
= 3 * 4* R1/ R1
= 12 A
Answer E
24 Jun 2015, 05:43
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The amount of electrical current that flows through a wire is inversely proportional to the resistance in that wire. If a wire currently carries 4 amperes of electrical current, but the resistance is then cut to one-third of its original value, how many of electrical current will flow through the wire?
A. 3/4 amperes
B. 4/3 amperes
C. 4 amperes
D. 8 amperes
E. 12 amperes
Solution:
Given that, I∝1/R
I1=4 amps, I2=?
R2=(1/3)R1
Substitute the values in the equation (I1/I2)=(R2/R1) -> (4/I2) = (R1/3R1)
-> I2 = 12. ANS E.
Thanks
Kudos Please
A. 3/4 amperes
B. 4/3 amperes
C. 4 amperes
D. 8 amperes
E. 12 amperes
Solution:
Given that, I∝1/R
I1=4 amps, I2=?
R2=(1/3)R1
Substitute the values in the equation (I1/I2)=(R2/R1) -> (4/I2) = (R1/3R1)
-> I2 = 12. ANS E.
Thanks
Kudos Please
