Answer:
0.232 mm.
Explanation:
Power, P = 96 W
Voltage, V = 120 V
Length of wire, l = 4.2 m
Let r be the radius of the wire.
Resistivity of nichrome, ρ = 1.5 x 10^-6 ohm metre
P = V^2 / R
R = 120 x 120 / 96 = 150 ohm
r = 1.16 x 10^-4 m
Diameter = 2 x r = 2 x 1.16 x 10^-4 = 2.32 x 10^-4 m = 0.232 mm
Thus, the diameter of the wire is 0.232 mm.
Answer:
The bulbs should be connected in parallel.
Explanation:
We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.
We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.
On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.
The current flowing through each bulb is given by
I = V/R
The voltage across each bulb is given by
V = IReq
Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by
Req = (R₁*R₂)/(R₁+R₂)
The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.
The best answer between the two options would be the second choice B) FALSE.
Given R2’s resistance and voltage we can find the current through it of 1.5 amps. Due to Kirchhoff’s junction rule the current going in must match the current going out so the current through them all is 1.5 amps. Using this we can find the voltage through R1 of 15v. Then we subtract V1+V2 from 120 to find that R3 has a voltage of 60v. Next we find that R3 has a resistance of 40 ohms