Question

Qn 1

A capacitor is attached to a battery with a terminal voltage of V. What happens to the capacitance of the capacitor if it is attached to a new battery with a terminal voltage of 2V, twice as large as the previous battery?

	The new capacitance is twice as large.
	The new capacitance is half as large.
	The new capacitance is the same as it was before.

Question 2

A capacitor is attached to a battery with a voltage of V, and a charge Q is pulled from the battery and stored on the capacitor. If a second identical capacitor is attached in parallel with the first, how much total charge has now been pulled from the battery?

	0.
	4Q.
	Q/2.
	2Q.

Question 3

A capacitor is attached to a battery with a terminal voltage of V. What happens to the charge on the the capacitor if it is attached to a new battery with a terminal voltage of 2V, twice as large as the previous battery?

Question 3 options:

	The new charge is twice as much as was stored before.
	The new charge is half as much as was stored before.
	The new charge is the same as was stored before.

Question 4

Two identical capacitors with capacitance C are connected together in series. How does the capacitance of the combination compare to C?

Question 4 options:

	It is larger than C.
	It is smaller than C.
	The capacitance of the combination is C.

Question 5

When current flowing through a resistor increases, the voltage drop across the resistor must be

Question 5 options:

	increasing.
	decreasing.
	remaining constant.

Question 6

A certain wire made from a conducting material with resistivity ρ{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>&#x3C1;</mi></math>"} has a certain length, L, and a certain diameter, D, giving it a resistance of R. If a second wire is made from a conductor with a resistivity that is twice as large, and the wire has a length of 2L and a diameter of 2D, how does the resistance of this wire compare to R?

Question 6 options:

	It has a resistance of R/4.
	It has a resistance of 4R.
	It has a resistance of 2R.
	It has a resistance of R/2.
	It has a resistance of R.

Question 7

A certain resistor is made of a semiconductor whose temperature coefficient of resistivity is negative. A constant voltage is applied to the resistor, but the resistor heats up significantly as current passes through the wire. What happens to the amount of current the resistor will carry?

Question 7 options:

	The current decreases.
	The current remains constant.
	The current increases.

Question 8

A capacitor is attached in series to a resistor with a resistance R. The combination is connected to a battery and allowed to charge. After a certain amount of time, T, the capacitor has charged to 60% of its maximum value. If we repeat the experiment, but replacing the resistor with a new resistor of resistance 2R, what happens to the time it takes for the capacitor to charge to 60% of its maximum value again?

Question 8 options:

	Need more information.
	It will take less time.
	It will take the same amount of time.
	It will take longer.

Question 9

A resistor with a resistance of 3 Ohms is connected in parallel with a resistance with a resistance of 6 Ohms. What is the equivalent resistance of this combination?

Question 9 options:

	2 Ohms.
	4.5 Ohms.
	9 Ohms.
	0.5 Ohms.

Question 10

In the diagram below, the current flowing through R1 is 0.3 Amps and the current flowing through R2 is 0.1 Amps. What is the current flowing through R3?

Question 10 options:

	0.4 Amps.
	Need more information.
	0.2 Amps.
	0.15 Amps.

Answer 1

(1) Capacitance will remin same as it dosent depend on the voltage of the battery.

(2) when capacitor are connected in parellal then overall capacitance of the system is C+C = 2C

Battery is same than from the equation

Q'= (2C)V

Q'=2Q

(3) Again using the same equation

new charge wil be = C(2V)

= 2Q

(4)Resultant capacitor is

1/C' = 1/C + 1/C

C'= C/2

hence new capacitance is smaller than c

(5) Voltage drop across a resistor is

V= IR

if I increases than V will also increase.

(6) the resistance of the wire is

  

and

So R will be

  

resistance for other wire will be

Cancelling out the 4 factor from numrator and denominator we get

hence R' = R

(7) If  temperature coefficient of resistivity is negative then resistance will decrease with rise in temprature. As the multiplication of resistance and current is constant (voltage), So current must increase in the circuit

(8) Charging of capacitor depends on the time constant of the circuit. initial it was

t= RC

but after replacing the resistor it will be

t'=2RC

t'=2t

so it will tke longer for it to charge now

(9) equivalent resistance is

1/R = 1/3 +1/6

1/R = 3/6

R=2 ohm

(10) Need Daigram for this , which is missing.