A charge of #8 C# passes through a circuit every #2 s#. If the circuit can generate #3 W# of power, what is the circuit's resistance?

Question

A charge of #8 C# passes through a circuit every #2 s#.  If the circuit can generate #3 W# of power, what is the circuit's resistance?

Isabella Alvarez · Accepted Answer

undefined To find the resistance of the circuit, we can use the formula for power: Power (P) = Current (I) * Voltage (V). Since power is given as 3 W and the current is equal to charge divided by time (I = Q/t), we can rearrange the formula to solve for resistance (R), which is equal to voltage divided by current (V/I).

First, let's find the current (I):
I = Q/t
I = 8 C / 2 s
I = 4 A

Now, let's use the power formula to find the voltage (V):
P = I * V
3 W = 4 A * V
V = 3 W / 4 A
V = 0.75 V

Finally, let's find the resistance (R):
R = V / I
R = 0.75 V / 4 A
R = 0.1875 ohms

Therefore, the circuit's resistance is 0.1875 ohms.

Isabella Alvarez · Answer

I found: #0.2Omega#

You know that current #I# is the rate of passing of charge #Q# through a section of a conductor with time #t# or:

#I=Q/t#

in your case: #I=8/2=4A#

Power #P# is the rate of transformation of energy with time that, in case of electrical circuits, can be written (using also Ohm's Law: #V=RI#):

#P=VI=V^2/R=I^2R#

using the last one:

#P=I^2R# and rearranging:

#R=P/I^2=3/4^2=3/16=0.1875~~0.2Omega#