MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 4

In a potentiometer wire experiment the emf of a battery in the primary circuit is 20V and its
internal resistance is 5

$\Omega$ . There is a resistance box in series with the battery and the potentiometer
wire, whose resistance can be varied from 120

$\Omega$ to 170

$\Omega$ . The resistance of the potentiometer wire is 75

$\Omega$ . Find the potential differences using a potentiometer.

0%
5V
0%
6V
0%
7.5V
0%
8V

A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths

$l_{1}$ cm and

$l_{2}$ cm respectively. The magnitudes, of the resistors R and X, in ohms, are then, equal respectively to:

0%
$k(l_{2}-l_{1})$ and $kl_{2}$
0%
$kl_{1}$ and $k(l_{2}-l_{1})$
0%
$k(l_{2}-l_{1})$ and $kl_{1}$
0%
$kl_{1}$ and $kl_{2}$

n identical cells are joined in series with its two cells A and B in the loop with reversed polarities.
EMF of each shell is E and internal resistance r. Potential difference across cell A or B is:

0%
$\dfrac{2E}{n}$
0%
$2E|1-\dfrac{1}{n}|$
0%
$\dfrac{4E}{n}$
0%
$2E|1-\dfrac{2}{n}|$

Explanation

Number of cells

$=n$

EMF of each cell

$=E$

Internal resistance of each cell,

$= r$

All the cells are in series.

Thus,

Total internal resistance,

$R=nr$

Total EMF,

$e=nE$

As the two cells are connected in reverse polarity, these two cells will cancel-out the contribution of other two cells in the loop.

Then, the net EMF across the circuit will be

$E_{eq}=nE-4E$

The current

$I$ in the circuit

$=\dfrac{nE-4E}{nr}$

Voltage across the opposite connected batteries,

$V=E-Ir$

$V=E-\dfrac{E(n-4)}{nr}\times r$

$V=\dfrac{4E}{n}$

Therefore, the voltage across the battery A or B will be

$\dfrac{4E}{n}$ .

Which of the following statement is true regarding the Kirchoffs laws?

0%
The junction rule is a statement of conservation of energy and the loop rule is a statement of conservation of charge
0%
The junction rule as well as the loop rule are statements of conservation of charge.
0%
The junction rule as well as the loop rule are statements of conservation of energy.
0%
The junction rule is a statement of conservation of charge and the loop rule is a statement of conservation of energy.

The resistance of a metal increases with the increase of temperature due to :

0%
Number of electrons
0%
Velocity of electrons
0%
Scattering of electrons with core ions
0%
Thermal motion of core ions

Observe the figure given below.
Find the current passing through

$6\Omega$ resistor.

0%
0.72 A
0%
0.80 A
0%
0.48 A
0%
Cannot be said

Explanation

Let the equivalent resistance between

$E$ and

$F$ consisting of

$R_1=6 \Omega$ and

$R_2=4 \Omega$ in parallel be

$R_{eq}.$

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{R_1+R_2}{R_1R_2}$

$\Rightarrow R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{(6)(4)}{6+4}=2.4 \Omega$

Let the voltage across

$E$ and

$F$ be

$V_{EF}.$

Applying Ohm's law for

$R_{eq}$

$V_{EF}=IR_{eq}$

$V_{EF}=(1.2 \ A) \times (2.4 \ \Omega)=2.88 \ V$

The voltage

$V_{EF}=2.88 \ V$ also appears across

$R_1$ and

$R_2$ .

$\therefore V_{EF}=V_{BC}=V_{AD}=2.88 \ V$

Let the current through

$6\Omega$ resistance be

$I_{BC}.$

Applying Ohm's Law for resistance

$R_1=6 \Omega$

$V_{BC}=I_{BC}R_1$

$\Rightarrow I_{BC}=\dfrac{V_{BC}}{R_1}=\dfrac{2.88}{6}=0.48 \ A$

Kilowatt-hour is the unit of :

0%
potential difference.
0%
electric power.
0%
electrical energy.
0%
charge.

Explanation

Kilowatt-hour $(KWh)$ is unit of Electrical Energy.

We know that,

$\text{Power} \times \text{time} = \text{Energy}$

$\Rightarrow\ \ \text{Kilo-Watt} \times \text{Hour} = \text{Energy}$

$1 KWh$ is electrical energy consumed when an electrical equipment of power $1\ KiloWatt$ is operated for $1\ Hour$ .
It is also used as commertial unit of energy in households/Buildings/Factories etc where 1 Unit of electricity is equivalent of 1KW-h of energy.

Hence, Option C is correct.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

You have a large supply of light bulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in series to the previous bulbs then

0%
Brightness of the bulbs will increase
0%
Current through the bulbs will increase
0%
Power transferred from the battery will decrease
0%
Lifetime of the battery will decrease

Find the currents

$I_1$ and

$I_2$ for the circuit shown in given figure.

0%
$I_1=-6 A$ and $I_2=9 A$
0%
$I_1=6 A$ and $I_2=9 A$
0%
$I_1=-6 A$ and $I_2=-9 A$
0%
$I_1=-9 A$ and $I_2=6 A$

Explanation

Consider node

$a$ , using Kirchhoff's current law

(using

$+$ for

$I_{in}$

$-$ for

$I_{out}$ )

$\sum I_{at\, a\, node} = 0$

$-3+2+I_{ab} = 0$

$I_{ab} = 1 A$

Consider node

$b$ ,

$I_{ba}+7+I_1 = 0$ (

$I_{ab} = -I_{ba}$ )

$-1-7 +I_1=0$

$I_1 = -6A$

Consider node

$d$ ,

$I_{dc} +4-1 = 0$

$I_{dc} = -3 A$

Consider node

$c$ ,

$-I_2+6+I_{cd} = 0$

$-I_2+6+3 = 0$

$I_2 = 9A$

$option\, A\, is\, correct$

$1kWh=$ _________?

0%
$3600000\ J$
0%
$10000\ J$
0%
$4.2\ J$
0%
$25000\ J$

Explanation

Kilowatt hour is the unit of energy to measure the amount of electricity used in an hour

$1\ kW$ in

$1\ hr$ .

Hence,

$1\ kWh = 1 kW \times 1\ hr$

$= 10^3 W \times 3600\ s$

$= 3600000\ J$

Kilowatt is the unit of electrical _______ but kilowatt-hour is the unit of electrical _______.

0%
Power, Energy
0%
Power, Electric potential
0%
Power, Current
0%
Energy, Power

Explanation

In SI system, unit of power is Watt (W). Kilowatt is equal to one thousand (

$10^3$ ) watts. The kilowatt-hour is a unit of energy and is equal to 1,000 watt-hours. If the energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time in hours.

What is one unit of household electrical energy in J?

0%
$1000\ J$
0%
$3600\ J$
0%
$0.6 \times 10^3 J$
0%
$3.6 \times 10^6 J$

Explanation

One unit of household is

$1\ kwh$

$kwh\ =$ kilo watt hour

$1\ \text{unit}$

$= 1\ kWh$

$\Rightarrow 1\ \text{unit} =10^3 W \times 3600\ s$

$\Rightarrow 1\ \text{unit} = 3.6 \times 10^6 J$

A negatively charged rod is brought close to two metal spheres which are in contact with each other, and the spheres are separated in the presence of the rod. Then

0%
The sphere close to the rod acquires a negative charge and the other sphere acquires a positive charge
0%
The sphere close to the rod acquires a positive charge and the other sphere acquires a negative charge
0%
Both the spheres will acquire positive charge
0%
Both the spheres will acquire negative charge

Explanation

Given that a negatively charged rod is brought closer to two metal spheres which are in contact with each other as shown:
So, free electrons from rod will be attracted towards the metal spheres. These electrons gets collected on the farther surface of sphere

$B$ and

$A$ spheres becomes short of electrons. So, when the two spheres get separated in the presence of rod, then sphere

$A$ (which is nearer to rod) acquires a positive charge and sphere

$B$ acquires a negative charge.

Two non-ideal batteries are connected in series. Consider the following statements:
(A) The equivalent emf is larger than either of the two emfs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances

0%
Each of A and B is correct
0%
A is correct but B is wrong
0%
B is correct but A is wrong
0%
Each of A and B is wrong

Explanation

When two non-ideal batteries are connected in series, their equivalent emf is

$E_{eq} = E_1 + E_2$
Hence, the equivalent emf is larger than either of the two emfs.
Also, their equivalent internal resistance is

$r_{eq} = r_1 + r_2$
Hence, the equivalent resistance is larger than either of the two internal resistances.
Therefore, statement A is correct but B is wrong.

Every atom makes one free electron in copper. If

$1.1$ ampere current is flowing in the wire of copper having

$1$ mm diameter, then the drift velocity (approx.) will be (Density of copper =

$9 \times 10^3\, kg m^{-3}$ and atomic weight =

$63$ )

0%
$0.1$ mm/s
0%
$0.2$ mm/s
0%
$0.3$ mm/s
0%
$2$ mm/s

The value of current

$i$ in the given circuit is:

0%
Zero
0%
5 Amp.
0%
7 Amp.
0%
11 Amp.

Explanation

$\textbf{Hint:}$

Kirchhoff's Current Laws states that net current coming towards a junction is zero.

$\textbf{Step 1: Note the values of current coming to and leaving the junction.}$

Current coming to junction is

$= 5+7+2 = 14A$

Current leaving the junction is

$= (3+i)A$

$\textbf{Step 2: Use Kirchhoff's Current Law and find the value of $$i$$.}$

Kirchhoff's First Law is also known as Kirchhoff's Current Law. It says that current coming towards junction is equal to the current leaving the junction.

By using KCL we have

$\text{Current coming to junction} = \text {Current leaving the junction}$

$\Rightarrow 14 = 3+i$

$\Rightarrow i=11A$

Thus, the value of $i$ is $11A$ .

Option D is correct.

When sphere C is brought in contact with the first sphere then find the charge left on sphere A and sphere C respectively.

0%
$6.5\times 10^-7C, 0C$
0%
$3.25\times 10^{-7}C, 3.25\times 10^{-7}C$
0%
$3.25\times 10^{-7}C, 6.5\times 10^{-7}C$
0%
$0C, 6.5\times 10^{-7}C$

Explanation

The original charge on the sphere A or B,

$q_1=q_2=6.5\times 10^{-7}C$
The distance between the two sphere,

$r=0.05m$
Since all the spheres are of same size, they will possess equal charges on being brought in contact. When uncharged sphere

$C(q_3=0)$ is brought in contact with A, charge left on the sphere A,

$q_1=\dfrac {q_1+q_3}{2}=\dfrac {6.5\times 10^{-7}+0}{2}$

$=3.25\times 10^{-7}C$
and charge on the sphere C,

$q_3=3.25\times 10^{-7}C$

The arm PQ can revolve with uniform speed continuously about P round the circular uniform potentiometer track XYZ. The voltage between RS will vary with respect to time :

0%
Sinusoidally
0%
Linearly
0%
Rectangularly
0%
Like saw tooth

Explanation

As the voltage between

$RS$

$(V) \propto$ Resistance between

$P$ and

$Z$

Arm

$PQ$ is revolving uniformly so resistance between

$P$ and

$Z$ is changing uniformly

Let resistance

$XZ = R$

So we can write resistance between

$P$ and

$Z$ as function of time :

$r = R - kt$

where

$k$ is some constant and

$r$ is resistance between

$P$ and

$Z$ which is a equation of decreasing straight line

As arm comes to point

$Z$ resistance

$r = 0$ and again it starts from

$R$ as it comes to point

$X$ .

There is a fixed potential difference between the two ends of a potentiometer. Two cells are connected in series in such a way that in one arrangement they help each other where as in the second arrangement they oppose each other. The balance point for these two combinations is obtained at 120 cm and 60 cm length respectively. The ratio of the emf of the cell is:

0%
2 :1
0%
3 :1
0%
1 : 1
0%
4: 1

Explanation

$E=\dfrac{x_1+x_2}{x_2}=\dfrac{120+60}{60}=3:1$

If a resistance

$5\Omega$ is connected in the left gap of a meter bridge and

$15\Omega$ in the other gap then position of balancing point is

0%
10 cm
0%
20 cm
0%
25 cm
0%
75 cm

Explanation

Consider the figure , it is given that

$R=5 \ ohm$ and

$S= 15 \ ohm$
Now, at the balance point the circuit will be like a Wheatstone bridge .
That is,

$R/S = l_{1}/ (100 - l_{1})$

$5/15 = l_{1}/ (100 - l_{1})$

$4*l_{1} = 100$

$l_{1} = 25cm$

1 kWh

$=$

0%
$36 \times 10^6$ watt-second
0%
$36 \times 10^5$ watt-second
0%
$36 \times 10^7$ watt-second
0%
$36 \times 10^8$ watt-second

A current of

$3\ A$ flows through a resistance when it is connected across a 12V source. What should the current be if the resistance is increased by

$4\ \Omega$ and connected across the same voltage source?

0%
$1.5\ A$
0%
4 A
0%
1.6 A
0%
24 A

Explanation

$\underline{\text{First Case:}}$

Voltage

$V_1=12\ V$

Current

$I=3\ A$

Resistance =

$R$

From Ohm's law:

$12=3\times R$

$\therefore R=\dfrac{12}{3} = 4\ \Omega$

$\underline{\text{Second Case:}}$

Voltage

$V_2=12\ V$

Resistance =

$R+4=8\ \Omega$

From Ohm's law:

$12=I'\times 8$

$\therefore I'=\dfrac{12}{8}=1.5\ A$

Which one of the following is different from the others?

0%
Joule
0%
Kilowatt hour
0%
Erg
0%
Watt

The given figure shows a network of currents. The current

$i$ is

0%
23 A
0%
22 A
0%
20 A
0%
25 A

Explanation

Here, consider the above part of circuit as shown in figure and applying Kirchhoff's current law we obtain

$I_1 + 5 -I_2 = 0$

$I_1 =I_2 - 5$ ..................................................(1)

$I_2 + 15 - I_3 = 0$

$I_2 = I_3 -15$ ..........................................(2)

$I_3 + 3 -I_4 = 0$

$I_3 = I_4 - 3$ ..........................................(3)

and

$I_4 - i - I_1 = 0$

$I_4 = I_1 + i$ ...............................................(4)

now using values from equations(2), (3) and (4) into (1)we get

$I_1 = I_1+i-23$

$\therefore i=23$ A

To measure a small resistance

$\sim 10^{-5}\Omega$ , one should used

0%
Wheatstone bridge
0%
Postoffice box
0%
Wein's bridge
0%
Carrey Foster bridge

Which device is used to measure the potential difference between two points of a conductor in the laboratory?

0%
Voltameter
0%
Ammeter
0%
Potentiometer
0%
Galvanometer

In a potentiometer wire experiment, the emf of a battery in the primary circuit is

$20V$ and its internal resistance is

$5\space\Omega$ . There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from

$120\space\Omega$ to

$170\space\Omega$ . Resistance of the potentiometer wire is

$75\space\Omega$ . The following potential differences can be measured using this potentiometer

0%
$5V$
0%
$6V$
0%
$7V$
0%
$8V$

Explanation

Here,

$E= 20$ V

$, r = 5$ ohm,

$R_l =75$ ohm and

$R$ varies from 120 ohm to 170 ohm.
Now, Current when

$R = 120$ ohm is

$I = \dfrac {E}{R+r+R_l}$

$I = \dfrac {E}{120+5+75}$

$I = 0.05$ A
Hence the voltage that can be measured is

$V = 0.05 \times120 = 6$ V
and when

$R= 170$ ohm, the current is

$I = \dfrac {20}{170+5+75} = 0.08A$

Hence the voltage that can be measured is

$V = 0.08 \times 170 = 13.6$ V
Hence potential difference from 6V to 13.6 V can be measured using potentiometer.

Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are

$V_1, V_2, V_3$

0%
$V_1=V_2$
0%
$V_1\neq V_2$
0%
$V_1+V_2=V_3$
0%
$V_1+V_2 > V_3$

Explanation

Here, three voltmeters are joined as shown in given figure.
We know,

$V = IR$
Also, all voltmeters have different resistances, then for same current

$V \propto R$
Hence,

$V_1 \propto R_1$
similarly,

$V_2 \propto R_2$ and

$V_3 \propto R_3$

$\Rightarrow V_1 \neq V_2$
When some potential difference is applied the current will divide into two branches i.e. say

$I_1$ in one having

$V_1$ and

$V_2$ in series and say

$I_2$ in other having

$V_3$ parallel to both of them.
Now, applying KVL to the loop

$I_1R_1 +I_1R_2- I_2V_3 = 0$

$V_1 + V_2 - V_3 = 0$

$V_1 + V_2 = V_3$

A brass disc and a carbon disc of same radius are assembled alternatively to make a cylindrical conductor. The resistance of the cylinder is independent of the temperature. The ratio of thickness of the brass disc to that of the carbon disc is

$\underline{\hspace{0.5in}}$ .

$\alpha$ is temperature coefficient of resistance & Neglect linear expansion

0%
$\displaystyle\frac{\alpha_C\rho_C}{\alpha_B\rho_B}$
0%
$\displaystyle\frac{\alpha_C\rho_B}{\alpha_B\rho_C}$
0%
$\displaystyle\frac{\alpha_B\rho_C}{\alpha_C\rho_B}$
0%
$\displaystyle\frac{\alpha_B\rho_B}{\alpha_C\rho_C}$

Explanation

Given the resistance is independent of temperature .
Hence ,

$\dfrac{\rho _B l _B}{A} ( \alpha _ B \Delta T ) = \dfrac{\rho _C l _C}{A} ( \alpha _ C \Delta T )$

$\therefore \dfrac{l_B}{l_C} = \dfrac{\rho _C \alpha _C}{\rho _B \alpha _B}$

Calculate the value of current

$I_2$ in the section of networks shown in figure.

0%
13 A
0%
7 A
0%
10 A
0%
3 A

Explanation

Applying KCL to the junction at point A, we get

$15 -8-I_2 = 0$

$\therefore I_2 = 7A$

Calculate the value of current

$I_4$ in the section of networks shown in figure.

0%
13 A
0%
7 A
0%
10 A
0%
3 A

In a balanced wheat stone bridge, current in the galvanometer is zero. It remains zero when

$[1]$ . battery emf is increased

$[2]$ . all resistances are increased by 10 ohms

$[3]$ . all resistances are made five times

$[4]$ . the battery and the galvanometer are interchanged

0%
only $[1]$ is correct
0%
$[1], [2]$ and $[3]$ are correct
0%
$[1], [3]$ and $[4]$ are correct
0%
$[1]$ and $[3]$ are correct

Explanation

The balancing condition for Wheatstone's bridge is,

$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$
This condition is independent of emf of cell, hence if battery emf is increased the bridge will remain balanced means no current will flow through the galvanometer. Also,if resistances are increased five times, we get

$\dfrac {5R_1}{5R_2} = \dfrac {5R_3}{5R_4}$

$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$

Hence condition remains the same.
If galvanometer and cell are interchanged,

$\dfrac {R_3}{R_4} = \dfrac {R_1}{R_2}$

$\dfrac {R_1}{R_2} = \dfrac {R_3}{R_4}$

Hence condition remains the same.
If resistances are increased by 10 ohms, we get

$\dfrac {R_1+10}{R_2+10} = \dfrac {R_3+10}{R_4+10}$

$( R_1+10)(R_4+10) = ( R_3+10)(R_2+10)$

$R_1R_4 +10(R_1+R_4) +100 = R_2R_3 +10(R_2+R_3) +100$

This will not yield the balanced condition for bridge.

Three ammeters A, B and C of resistances

$R_A, R_B$ and

$R_C$ respectively are joined as shown. When some potential difference is applied across the terminals

$T_1$ and

$T_2$ , their readings are

$I_A, I_B$ and

$I_C$ respectively

0%
$I_A=I_B$
0%
$I_AR_A+I_BR_B=I_CR_C$
0%
$\frac {I_A}{I_C}=\frac {R_C}{R_A}$
0%
$\frac {I_B}{I_C}=\frac {R_C}{R_A+R_B}$

Explanation

According to the given data, ammeters A and B are connected in series hence will read the same current through them i.e.

$I_A = I_B$ ..........(1)
Now, applying KVL we get,

$I_AR_A + I_BR_B - I_CR_C = 0$

$I_AR_A + I_BR_B = I_CR_C$ ............(2)
Using equation (1) in (2) we get

$I_BR_A + I_BR_B = I_CR_C$

$I_B(R_A + R_B) = I_CR_C$

$\frac{I_B}{I_C} = \frac{R_C}{(R_A + R_B)}$

$\frac{I_A}{I_C} = \frac{R_C}{(R_A + R_B)}$ ........from(1)

Calculate the value of current

$I_3$ in the section of networks shown in figure.

0%
13 A
0%
7 A
0%
10 A
0%
3 A

Two cells of same emf are connected in series. Their internal resistances are

$r_1$ and

$r_2$ respectively and

$r_1 > r_2$ . When this combination is connected to an external resistance R then the potential difference between the terminals of first cell becomes zero. In this condition the value of R will bw

0%
$\frac {r_1-r_2}{2}$
0%
$\frac {r_1+r_2}{2}$
0%
$r_1-r_2$
0%
$r_1+r_2$

Explanation

Let emf of each cell is

$E$ .
As they are connected in series so current in circuit is

$I=\dfrac{E+E}{r_1+r_2+R}=\dfrac{2E}{r_1+r_2+R}$

Potential across terminal of first cell is

$V_1=E-Ir_1=E-\dfrac{2Er_1}{r_1+r_2+R}$

$V_1=0 \Rightarrow E-\dfrac{2Er_1}{r_1+r_2+R}=0$

$r_1+r_2+R-2r_1=0$

$R=r_1-r_2$

In an experiment of Wheatstone bridge, if the positions of cells and galvanometer are interchanged, then the balance points will

0%
change
0%
remain unchanged
0%
depend on the internal resistance of the cell and resistance of the galvanometer
0%
none of these

Explanation

When Wheatstone bridge is balanced,

$\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4} or \dfrac{R_1}{R_3}=\dfrac{R_2}{R_4}$

If the galvanometer is replaced with a cell in balanced Wheatstone bridge, the condition for balanced bridge will be unchanged. i.e

$\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}$ . In balanced condition no current pass through the galvanometer. Thus, the null points will remian same.

The reading of the ammeter is:

0%
$\dfrac { 1 }{ 8 }\ A$
0%
$\dfrac { 3 }{ 4 }\ A$
0%
$\dfrac { 1 }{ 2 } \ A$
0%
$2\ A$

Consider the following two statements:
(A) Kirchhoff's Junction Law follows from conservation of charge.
(B) Kirchhoff's Loop Law follows from conservative nature of electric field.

0%
Both A and B are correct
0%
A is correct but B is wrong
0%
A is correct but A is wrong
0%
Both A and B are wrong

Which of the following statements is/are correct for potentiometer circuit?

0%
Sensitivity doesn't depend on the length of the potentiometer wire.
0%
Sensitivity is inversely proportional to potential difference applied across the potentiometer wire.
0%
Accuracy of potentiometer can be increased, only by increasing length of the wire.
0%
Range is independent of potential difference applied across the potentiometer wire.

Explanation

$\text{Sensitivity} \$

$\propto \dfrac{1}{\text{length of potentiometer wire}}$

$\propto \text{potential difference across potentiometer wire}$

On increasing length of potentiometer wire, the least count decreases, hence the accuracy increases. Hence the correct option is option C.

If I current is flowing in a potentiometer wire of length

$L$ and resistance

$R$ , then potential gradient will be

0%
$\dfrac {IR}{L}$
0%
$IRL$
0%
$\dfrac {RL}{I}$
0%
$\dfrac {IL}{R}$

Explanation

The potential is the fall of potential per unit length of potentiometer wire, hence

$k = \dfrac{V}{L}$

$k = \dfrac{IR}{L}$

What is the current I for the circuit shown in the figure?

0%
$3A$
0%
$1A$
0%
$-5A$
0%
$5A$

Explanation

total current leaving the circuit = total current entering the circuit

$I + 4 = -3 + 2$

$I= -5 A$

$1^{\circ}C$ rise in temperature is observed in a conductor by passing a certain current. If the current is doubled, then the rise in temperature is approximately

0%
$2.5^{\circ}C$
0%
$4^{\circ}C$
0%
$2^{\circ}C$
0%
$1^{\circ}C$

Explanation

We know that heat

$H=i^2Rt$ so

$H\propto i^2$
So when en current i is double, heat will four times increase.
As heat is proportional to change in temperature i.e,

$H\propto \Delta T$ so the rise in temperature will be

$4^oC$ .

What is the p.d. across the terminals

$(V_T)$ of a cell with emf E for open circuit ?

0%
$V_T< E$
0%
$V_T>E$
0%
$V_T=0$
0%
$V_T=E$

Choose the correct option.

0%
$kW=$ unit of power and $kWh =$ unit of energy.
0%
$kW=$ unit of energy and $kWh =$ unit of power.
0%
$kW$ and $kWh$ are both of unit of energy.
0%
$kW$ and $kWh$ are both of unit of power.

Explanation

Electrical Power is measured in watts (

$W$ ), kilo-watts (

$kW$ ), mega-watts (

$MW$ ), and giga-watts (

$GW$ ).

Electrical Energy is measured in kilowatt-hours (

$kWh$ ), megawatt-hours, and sometimes (less commonly) in watt-hours or watt-seconds.

A cell of e.m.f.

$\varepsilon$ and internal resistance r is used to send current to an external resistance R. The total resistance of circuit,

0%
$\dfrac{Rr}{(R+r)}$
0%
$\dfrac{R+r}{2}$
0%
$Rr$
0%
$R+r$

Explanation

Let us consider the circuit as given in the diagram below.
Batteries and cells have an internal resistance (r) which is measures in ohms. When electricity flows round a circuit the internal resistance of the cell itself resists the flow of current and so thermal (heat) energy is wasted in the cell itself.

$\varepsilon =I(R+r)$
where,

$\varepsilon$ = electromotive force in volts, V
I = current in amperes, A
R = resistance of the load in the circuit in ohms,
r = internal resistance of the cell in ohms.
The above equation can be written as

$\varepsilon =IR+Ir$ .
Hence, the total resistance in the circuit is given as R+r.

Two resistors of resistance

$4\, \Omega$ and

$6\, \Omega$ are connected in parallel to a cell to draw

$0.5 A$ current from the cell. What is the current through resistor

$6\, \Omega$ ?

0%
$0.2 \ A$
0%
$0.5 \ A$
0%
$2.4 \ A$
0%
$1.2 \ A$

Explanation

Two resistors of resistance

$4 \ Ω$ and

$6 \ Ω$ are connected in parallel.

Hence, equivalent resistance in parallel is given by,

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

$\Rightarrow \dfrac{1}{R_{eq}} = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{10}{24}$

$\Rightarrow R_{eq} = \dfrac{24}{10}= 2.4 \ \Omega$

Using Ohm's law,

$V=iR$

$V=0.5\times2.4=1.2 \ V$

Since, value of voltage is same for objects connected in parallel, Hence, voltage drop across

$6 \ \Omega$ resistor is also

$1.2\ V$

Hence, using Ohm's law , Current through resistor

$6 \ Ω$ :

$1.2=i\times6$

$i=0.2 \ A$

2106174_174091_ans_900459579ded4f159b9945852d36b55a.png

The power supplied by the battery will be :

0%
15 W
0%
24 W
0%
3.6 W
0%
20 W

Explanation

In this case, the resistances 4 ohms and 6 ohms are connected in parallel.
The total resistance is

$\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 6 } =\dfrac { 1 }{ 0.416 } =2.4\quad ohms$ . The voltage is given as 6 V.
The power is calculated as follows.

$P=VI\quad =\quad \dfrac { { V }^{ 2 } }{ R } (as\quad I=\dfrac { V }{ R } )$ .
Therefore,

$P=\quad \dfrac { { 6 }^{ 2 } }{ 2.4 } \quad =\quad 15W$ .
Hence, the power supplied by the battery is 15 W.

In an experiment of verification of Ohm's law, following observations are obtained:

Potential difference V (in volt)	0.5	1.0	1.5	2.0	2.5
Current I (in ampere)	0.2	0.4	0.6	0.8	1.0

What will be the potential difference V when the current I is 0.5 A ?

0%
$1.25 V.$
0%
$0.5 V$
0%
$2 V$
0%
$1 V$

The e.m.f. of the cell is

0%
$1.2 V$
0%
$4.8 V$
0%
$1.8 V$
0%
$2 V$

Explanation

In the first case, there are 2 resistors of 2 ohms each connected in parallel.

The combined resistance is given as

$\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } =\dfrac { 1 }{ 1 } =1\ ohms$ .
In the second case, there are 2 resistors of 2 ohms each connected in series. The combined resistance is given as

$2+2 = 4$

$ohms$ .

The current in the circuit is given by the formula

$I=\dfrac { \varepsilon }{ R+r }$ where,

$\varepsilon$ - emf of the cell
R- external resistance
r- internal resistance of the cell.

The current in the 1 st case is given as

$1.2=\dfrac { \varepsilon }{ 1+r } ,\quad that\quad is,\quad 1.2+1.2r=\varepsilon$ - eqn - 1
The current in the 2nd case is given as

$0.4=\dfrac { \varepsilon }{ 4+r } ,\quad that\quad is,\quad 1.6+0.4r=\varepsilon$ - eqn - 2

Solving eqn 1 and eqn 2 for r and

$\varepsilon$ , we get,

$r = 0.5$ ohms and

$\varepsilon$

$= 1.8$ volts

Hence, the emf of the cell is given as 1.8 volts.

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 4 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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