"Nearly every man who develops an idea works
it up to the point where it looks impossible, and then he gets discouraged.
That's not the place to become discouraged."

Today we will discuss a few multiple choice practice questions on electric circuits. Questions in this section were discussed on earlier occasions. You may click on the label ‘electric circuits’ below this post to access all posts in this section. Here are the questions with their solution:

(1) The voltmeter in
the shown shown has a resistance of. The internal resistance of
the 10 V battery is negligible. What is the reading of the voltmeter?

(a) 2 V

(b) 3.3 V

(c) 4 V

(d) 5V

(e) 6 V

The voltmeter of
resistance 100 KΩ is connected
across a 100 KΩ resistor and
these two give a parallel combined resistance value of 50 KΩ. The total resistance in series with the
battery is 100 KΩ + 100 KΩ + 50 KΩ
= 250 KΩ.

The current in the series circuit is 10 V/250 KΩ and hence the potential difference across the
parallel combination of the voltmeter and the 100 KΩ resistor is (10 V/250 KΩ) × 50 KΩ = 2 V [Option
(a)].

[We retained the resistances in KΩ itself in the above expressions so that we could obtain the
potential difference in

*volts*in a convenient manner].
(2) The adjoining figure
shows junctions J

_{1}and J_{2}of an electric circuit. Currents at these junctions sufficient for arriving at the answer are indicated in the figure. What is the value of the current*I*?
(b) 3 A

(c) 4 A

(d) 8 A

(e) 10 A

In accordance with
Kirchoff’s current law (KCL), the current

*I*_{1}flowing*outwards*from junction J_{1 }towards junction J_{2}is 5 A – 2 A = 3 A.
[Algebraic sum of
currents at a junction is zero according to KCL:

5 A – 2 A +

*I*_{1 }= 0
Therefore,

*I*_{1 }= – 3 A, the negative sign showing that the current flows outwards from the junction].
Total current
flowing towards junction J

_{2}is 3 A + 5 A = 8 A.
Since a current of
4 A is shown as

*flowing out*from junction J_{2}, the remaining current*I*_{2 }that has to*flow out*must be 4 A [Option (c)].
(3) A 2 μF capacitor is
connected in series with a 1 μF capacitor and a 60 V power supply. The potential difference
across the 2 μF capacitor is

(a) 60 V

(b) 40 V

(c) 30 V

(d) 20 V

(e) zero

The effective capacitance connected across the
power supply is the series combined value of the 2 μF and 1 μF
capacitors which is (

^{2}/_{3}) μF
[If capacitors C

_{1}and C_{2}are connected in series, their effective capacitance C is given by the reciprocal relation, 1/C = 1/ C_{1 }+ 1/C_{2}so that C = C_{1}C_{2}/( C_{1}+ C_{2})]
The charge

*Q*supplied by the power supply is given by*Q = CV =*(

^{2}/

_{3}) × 60 = 40 μC

[We obtain the charge in micro coulomb since the capacitance is in μF].

The capacitors hold the

*same*charge in series connection. Therefore, the voltage across the 2 μF
capacitor is

*Q/C*_{1}= 40 μC/2 μF = 20 volt [Option (d)].
[You will be able to work out this problem in no time if you remember
that the voltages are distributed among series connected capacitors in

*inverse proportion*to their capacitances].

*The following questions are specifically meant for AP Physics C aspirants:*
(4) The resistance

*R*of any uniform wire of length*L*and cross section area*A*is related to the resistivity (or, specific resistance)*ρ*of its material as*R =*

*ρL/A*

Suppose you have

*n*wires of the same length*L*and the same area of cross section*A*made of materials of resistivities*ρ*, 2*ρ*, 3*ρ*, 4*ρ*, ……*nρ*respectively. If they are connected in series, what will be the resistivity of the material of the combined wire?
(a)

*n ρ*
(b)

*n*(n+1)*ρ*
(c)

*n ρ/*2
(d) (

*n*+1)*ρ*/2
(e)

*n*(*n*+1)*ρ*/2
The resistances of
the 1

^{st}, 2^{nd}, 3^{rd}, 4^{th}, ……*n*^{th }wires are respectively given by*R*_{1}=*ρL/A*,*R*_{2}= 2*ρL/A*,*R*_{3}= 3*ρL/A*,*R*_{4}= 4*ρL/A*, …….*R*=_{n}*n**ρL/A*.
The total resistance

*R*of the series combination is given by*R =*

*R*

_{1}+

*R*

_{2}+

*R*

_{3 }+

*R*

_{4}+ ……+

*R*

_{n}
Or,

*R =*(*ρL/A*)(1 + 2 + 3 + 4 + ……*n*), on substituting for*R*_{1},*R*_{2},*R*_{3}etc.
This gives

*R =*(*ρL/A*)[n(n + 1)/2] =*n*(*n*+ 1)*ρL/*2*A*
The length of the compound wire is

*nL*and its area of cross section is*A*. If the resistivity is*ρ*its resistance can be written as*R =*

*ρnL/A*

Therefore

*ρ = RA/nL =*[*n*(*n*+ 1)*ρL/*2*A*](*A/nL*) = (*n*+1)*ρ*/2
[Even without writing all the above steps you
could have written the answer, arguing that the resistivity of the combined wire
must be the average value of the resitivities of the individual wires, since
the increments in resistivity occur in a regular manner. Therefore

*ρ =*(*ρ*+2*ρ*+ 3*+*4*ρ+*……+*nρ*)/*n =*(*n*+1)*ρ*/2]
(5) In the circuit
shown in the adjoining figure, the ammeter reading is zero when the switch S is
closed. If the batteries are of zero internal resistance, the value of
resistance X is

(b) 200 Ω

(c) 300 Ω

(d) 400 Ω

(e) 600 Ω

Since the current
through the ammeter is zero, the voltage drop across the series combination of
X and 600 Ω must be 4 V.

[The battery emf of 4 V and the p.d. across the series
combination of X and 600 Ω have tu be equal and
and in

*opposition*to attain the condition of zero current through the ammeter].
The 6 V battery
drives a current through the three resistors and 4 V is dropped across the
series combination of X and 600 Ω.
The remaining 2 V is dropped across the 400 Ω resistor.

Since the potential drop across the series
combination of X and 600 Ω is 4V, we have

X + 600 Ω = 400 Ω

Therefore X = 200 Ω

[If you want to use
Kirchoff’s laws and write down mathematical steps for solving the above
problem, here is how you will proceed:

For the loop containing the 4 V
battery, ammeter, X and 600 Ω,
we have

(I

_{1}+ I_{2})(X+600) = 4 where I_{1}and I_{2}are the currents supplied by the 4 volt and 6 volt batteries respectively.
Since I

_{1}= 0 we have
I

_{2}(X+600) = 4 …………..(i)
For the loop
containing the 6 V battery, 400 Ω, X and 600 Ω, we have

I

_{2}(400+X+600) = 6 ……..(ii)
Dividing Eq. (i) by Eq. (ii), we have

(X+600)
/(400+X+600) = 2/3

This gives X = 200 Ω]

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