Few weeks have been elapsed after my last post on this blog. I have been extremely busy during the last few weeks and now I am a bit relieved.

Today I’ll give you a few multiple choice practice questions on electric circuits. You will find a few earlier posts in this section elsewhere on this blog, which you can access by clicking on the label ‘direct current circuit’ below this post. The questions I give below are meant for checking your grasp of the fundamental principles:

(1) A battery of constant emf

*V*volt and negligible internal resistance is connected across a non uniform wire of nichrome having length*L*and resistance*R*. Pick out the correct statement from the following:
(a) The currents at all sections of the wire are not the same

(b) The potential drop per unit length of the wire is constant

(c) The current density along the wire is uniform

(d) The drift velocity of the charge carriers in the wire is non-uniform

(e) All the above statements are wrong

The expression for the electric current

*I*in the wire is*I = navq*where

*n*is the number of charge carriers per unit volume,

*a*is the area of cross section,

*v*is the drift velocity of the charge carriers and

*q*is the charge on each carrier.

[In conductors the charge carriers are electrons and

*q = e*, the electronic charge].
The current everywhere is the same and hence the statements (a), (b) and (c) are wrong. The only correct statement is (d) since the drift velocity as given from the above equation will be different, depending on the area of cross section of the wire.

(2) A cell of emf

*V*volt and internal resistance*r*ohm is connected across an external resistance 5*r*. The terminal voltage (terminal potential difference) of the cell is
(a)

*V/*6
(b) 5

*V/*6
(c) 6

*V/*5
(d)

*V/*5
(e)

*V/*4
Since the external resistance is across the cell, the terminal potential difference is the same as the potential difference across the external resistance.

The current through the external resistance is

*V/*(*r+*5*r*) =*V/*(6*r*).
The potential difference across the external resistance is [

*V/*(6*r*)] × 5*r*= 5*V/*6
(3) A current

*I*flows through a resistive network as shown in the adjoining figure. If the power dissipated in the 8 Ω resistor is 32 watt, what is the quantity of heat generated per second in the 1 Ω resistor?
(b) 4 J

(c) 8 J

(d) 32 J

(e) 64 J

Since the power dissipated in the 8 Ω resistor is 32 watt, we have

*I*

_{1}

^{2}×8 = 32 where

*I*

_{1}is the current through the 8 Ω resistor.

Therefore,

*I*_{1}= 2 A.
Resistors

*R*_{1}and*R*_{2}which make a total resistance of 12 Ω in the upper branch of the network carry the same current of 2 A. Since the total resistance of the lower branch (consisting of*R*_{3 }and*R*_{4}) of the network is 3 Ω only (which is a*quarter*of the resistance of the upper branch), the current through the lower branch is 4 times the current through the upper branch.
The current through the 1 Ω resistor is therefore equal to 8 A.

The heat generated per second in the 1 Ω resistor is 8

^{2}×1 = 64 joule [Option (e)].
(4) Four equal resistors connected in series across a battery of negligible internal resistance dissipates a total power of 200 milliwatt. If the parallel combination of these resistors is connected across the same battery, the total power dissipated in them will be

(a) 1.2 W

(b) 1.6 W

(c) 3.2 W

(d) 4.8 W

(e) 6.4 W

If the emf of the battery is

*V*volt and the value of each resistor is*R*ohm, we have, in the first case (when the series combination is across the battery)*V*

^{2}/4

*R*= 0.2 watt ………(i)

In the second case (when the parallel combination is across the battery), if the total power dissipated is

*P*, we have*V*

^{2}/(

*R/*4) =

*P*…………..(ii)

Dividing Eq (i) by Eq (ii), 1/16 = 0.2/

*P*so that*P*= 3.2 W.
The questions given above are meant for AP Physics B as well as AP Physics C aspirants. The following question is meant specifically for AP Physics C aspirants:

(5) The adjoining figure shows an infinite ladder network in which the resistors are of the same value, each equal to 1 Ω. The effective resistance between the points A and B is nearly

(a) infinite

(b) 2.333 Ω

(c) 2.505 Ω

(d) 2.732 Ω

(e) 2.999 Ω

Let us suppose that the effective resistance between the points A and B is R

_{e}. Since the ladder is infinitely long, we can add to it one more section containing three 1 Ω resistors as shown in the adjoining figure.
The new terminals are A

_{1}and B_{1}and the effective resistance of this modified infinite ladder between these terminals will still be R_{e}. Therefore we have
R

_{e}= 1 + [1×R_{e}/(1+R_{e})] + 1
[The second term on the right hand side of the above equation is the parallel combined value of 1 Ω and R

_{e}]
Or, R

_{e}+ R_{e}^{2}_{ }= 2 + 2R_{e}+ R_{e}
Thus R

_{e}^{2}– 2R_{e}– 2 = 0
This quadratic equation gives R

_{e}= [2**±**√(4+8)]/2
Or, R

_{e}= 1**±**√3
The effective resistance has to be

*positive*. So the answer is 1+√3 = 2.732 Ω, very nearly.
You will find similar multiple choice questions with solution here.