A current flows through the wire. The electric field in the copper section of th

Question

A current flows through the wire. The electric field in the copper section of the wire is E_Cu = 0.010 V/m. What is the electric field E_Al in the aluminum section of the wire? Data: sigma Cu = 6.0 times 10^7 Ohm^-1 m^-1 sigma Al = 3.5 times 10^7 Ohm^-1 m^-1 In the current shown in Fig, (2),describe what happens to the brightness of bulbs A and B when the switch is closed. Explain. Find the current through and the potential across each resistor in the circuit shown in Fig. (3) Find the rate that energy is supplied by the battery. Consider the circuit in Fig. (4) and assume the filament bulb is Ohm c with R = 50 Ohm. The capacitor is initially uncharged when the switch is closed. Make a sketch of the brightness of the bulb as a function of time. Clearly justify your graph including a discussion of Kirchoff's Voltage Law. A long time after the switch is closed, how much power is dissipated in the light bulb? After the switch has been closed for a long time it is reopened. Make a sketch of the brightness of the bulb as a function of time after the switch is opened. Clearly justify your graph. Reconsider parts (a) and (c). Which if either reaches it's final state faster and why?

Explanation / Answer

2.

initially if switch is open.

total resistance will be

Rt = Ra + Rb = Rab

it = V/Rab

Just after switch is closed, you can see that wire parallel to bulb B have zero resistance.

Now Req of bulb B and parallel zero resistance wire will be

1/Req = 1/0 + 1/Rb

1/Req = infinity

Req = 0

So combined resistance of bulb B and wire will be zero.

Total resistance of circuit will be

Rt = Ra + Req

Rt = Ra

Now you can see that

Rab >> Ra

So initially when switch is open current was low in the circuit in compare to when switch was closed

Power = V*i

Since after closing the switch current is increased in bulb A, so power in bulb A will also increase

Using parallel distribution of current in bulb B and zero resistance wire, current in bulb B will be zero, because whole current will pass through zero resistance wire.

Since after closing the switch current is decreased in bulb B, so power in bulb B will also decrease

Let me know if you have any doubt.

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