1. A copper cylinder is connected to an ideal battery. If the cylinder is heated
ID: 2256532 • Letter: 1
Question
1. A copper cylinder is connected to an ideal battery. If the cylinder is heated, how does the current through the copper cylinder change? A. Current Increases B. Current Decreases C. Current stays the same
2. A copper cylinder is connected to an ideal battery. If the cylinder is heated, how does the voltage drop across the the copper cylinder change? A. Voltage Drop increases B. Voltage Drop decreases C. Voltage drop stays the same
3. The plates of a capacitor are pushed together while remaining connected to the battery. How willl each of the following change as the plates are pushed together?
The charge on the plates Increase/Decrease/Remain the same
The potential drop across the plates Increase/Decrease/Remain the same
The electric field between the plates Increase/Decrease/Remain the same
The capacitance of the capacitor Increase/Decrease/Remain the same
4. A capacitor is charged by a battery. The battery is disconnected with the plates insulated so no charge escapes. How does the following change?
The charge on the plates Increase/Decrease/Remain the same
The potential drop across the plates Increase/Decrease/Remain the same
The electric field between the plates Increase/Decrease/Remain the same
The capacitance of the capacitor Increase/Decrease/Remain the same
The potential drop across the plates Increase/Decrease/Remain the same
The electric field between the plates Increase/Decrease/Remain the same
The capacitance of the capacitor Increase/Decrease/Remain the same
The potential drop across the plates Increase/Decrease/Remain the same
The electric field between the plates Increase/Decrease/Remain the same
The capacitance of the capacitor Increase/Decrease/Remain the same
Explanation / Answer
1)
resistance will increase as temperature increases..
so.. B) current decreases
2)
C) voltage drop stays same
3) Capacitance = Ae / d
so.. as d decreases.. C will increase
so.. voltage will remain same..
so.. charge = C*V
so.. charge will increase ..
The charge on the plates Increase
The potential drop across the plates Remain the same
electric field = voltage / d ... as as d decrease.. electric field increse
The electric field between the plates Increase
The capacitance of the capacitor Increase
4)
The charge on the plates Remain the same
The potential drop across the plates Remain the same
The electric field between the plates Remain the same
The capacitance of the capacitor Remain the same
5)
capaccitance is given by :
C = A * dielectric constant / d
The charge on the plates Decrease
The potential drop across the plates Remain the same
The electric field between the plates Decrease
The capacitance of the capacitor Decrease
6)
The charge on the plates Remain the same
The potential drop across the plates Increase
The electric field between the plates Increase
The capacitance of the capacitor Decrease
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