The current in the wires of a circuit is 60.0 milliAmps. If the voltage impressed across the ends of the circuit were halved (i. e., one-half of the original voltage) (with no change in its resistance), then its new current would be milliAmps

30 miliAmps

Explanation:

Step 1:

Obtaining an expression to solve the question. This is illustrated below:

From ohm's law,

V = IR

Were:

V is the voltage.

I is the current.

R is resistance.

From the question given, we were told that the resistance is constant. Therefore the above equation can be written as shown below:

V = IR

V/I = constant

V1/I1 = V2/I2

V1 is initial voltage.

V2 final voltage.

I1 is initial current.

I2 final current.

Step 2:

Data obtained from the question. This include the following:

Initial voltage (V1) = V

Initial current (I1) = 60 miliAmps

Final voltage (V2) = one-half of the original voltage = 1/2V = V/2

Final current (I2) =..?

Step 3:

Determination of the new current. This can be obtained as follow:

V1/I1 = V2/I2

V/60 = (V/2) / I2

Cross multiply to express in linear form

V x I2 = V/2 x 60

V x I2 = V x 30

Divide both side by V

I2 = (V x 30)/V

I2 = 30mA.

Therefore, the new current is 30miliAmps

this is poiselle's equation from fluid mechanic in physics. mech eng

it has to be b, about 97 percent.