A cell of e.m.f. $$\varepsilon$$ and internal resistance $$r$$ sends current $$1.0 A$$ when it is connected to an external resistance $$1.9\, \Omega$$. But it sends current $$0.5 A$$ when it is connected to an external resistance $$3.9\, \Omega$$. Calculate the values of $$\varepsilon$$ and $$r$$.

$$\varepsilon\, =\, 2.0\, V,\quad\, r\, =\, 0.1\, \Omega$$

$$\varepsilon\, =\, \, .1 V,\quad\, r\, =\, 2.0\, \Omega$$

$$\varepsilon\, =\, \, .2 V,\quad\, r\, =\, 1.0\, \Omega$$

In a typical Wheatstone network the resistance in cyclic order are $$A=10\:\Omega ,\:B=5\:\Omega ,\:C=4\: \Omega $$ and $$D=4\:\Omega $$ for the bridge to be balanced :-

$$5\:\Omega $$ should be connected in series with $$B$$

$$10\:\Omega $$ should be connected in parallel $$A$$

$$10\:\Omega $$ should be connected in series with $$A$$

$$5\:\Omega $$ should be connected in parallel with $$B$$

In a battery five dry cells each of 1.5 volt have internal resistance of $$0.2, 0.3, 0.4, 0.5$$ and $$0.6\Omega $$ are present in series. The battery is connected to a $$1 \Omega$$ resistance. Identify the correct statement (s) :

Current in the circuit will be $$2.5 A$$

Current in the circuit will be $$1.5 A$$

On short circuiting the battery $$3.75A$$ current will flow

MCQ Questions for Class 12 Medical Physics Current Electricity Quiz 5

A cell of e.m.f.

ε

$\varepsilon$ and internal resistance

r

$r$ sends current

1.0 A

$1.0 A$ when it is connected to an external resistance

1.9 Ω

$1.9\, \Omega$ . But it sends current

0.5 A

$0.5 A$ when it is connected to an external resistance

3.9 Ω

$3.9\, \Omega$ . Calculate the values of

ε

$\varepsilon$ and

r

$r$ .

0%
$\varepsilon\, =\, 2.0\, V,\quad\, r\, =\, 0.1\, \Omega$
0%
$\varepsilon\, =\, \, .1 V,\quad\, r\, =\, 2.0\, \Omega$
0%
$\varepsilon\, =\, \, .2 V,\quad\, r\, =\, 1.0\, \Omega$
0%
Data insufficient

Explanation

When the circuit is closed, the resulting current not only flows through the external circuit, but through the source (cell) itself. The cell have an internal resistance, which causes an internal voltage drop, slightly reducing the voltage across the terminals. The larger the current, the larger the internal voltage drop, and the lower the terminal voltage.
The current in the circuit is calculated from the formula

I = \frac{ε}{R + r}

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

ε

$\varepsilon$ - emf of the cell
R- external resistance
r- internal resistance of the cell.
When the cell is connected to the resistance of 1.9 ohms the current relation is given as
That is,

1.0 = \frac{e}{1.9 + r} . T h a t i s, a f t e r s i m p l i f i c a t i o n, e - r = 1.9 - e q n 1

$1.0=\dfrac { e }{ 1.9+r } .\quad That\quad is,\quad after\quad simplification,\quad e-r=1.9\quad -\quad eqn\quad 1$ .
When the cell is connected to the resistance of 3.9 ohms the current relation is given as
That is,

0.5 = \frac{e}{3.9 + r} . T h a t i s, a f t e r s i m p l i f i c a t i o n, e - 0.5 r = 1.95 - e q n 2

$0.5=\dfrac { e }{ 3.9+r } .\quad That\quad is,\quad after\quad simplification,\quad e-0.5r=1.95\quad -\quad eqn\quad 2$ .
Solving the linear eqn 1 and eqn 2 for the variables e and r, we get e=2.0 V.
Substituting the value of e in eqn 1, the variable r is determined to be 0.1 ohms.
Hence, the emf and the internal resistance of the cell are 2.0 V and 0.1 ohms respectively.

In our household applications, commercial unit of electricity is used. One unit is equal to:

0%
$1\ kW$
0%
$1\ kWh$
0%
$1\ kW/h$
0%
$1\ Wh$

What is the approximate resistance setting of a rheostat in which

650 m A

$650 \ mA$ of current flows with a

150 V

$150 \ V$ source?

0%
$230 \ \Omega$
0%
$56 \ \Omega$
0%
$15 \ \Omega$
0%
$13 \ \Omega$

Explanation

Given:

Current

I = 650 m A = 650 \times 10^{- 3} A

$I=650 \ mA=650 \times 10^{-3} \ A$

Voltage

V = 150 V

$V=150 \ V$

By Ohm's Law,

R = \frac{V}{I}

$R=\dfrac{V}{I}$

⟹ R = \frac{150}{650 \times 10^{- 3}}

$\implies R=\dfrac{150}{650\times 10^{-3}}$

⟹ R \approx 230 Ω

$\implies R\approx 230 \ \Omega$

Answer: Option (A)

The four wires from a larger circuit intersect at junction A a shown. What is the magnitude and direction of the current between points A and B?

0%
2 A from A to B
0%
2 A from B to A
0%
3 A from A to B
0%
3 A from B to A

Explanation

Kirchhoff's junction rule states that the algebraic sum of all currents into and out of branch point is zero i.e,

Σ I = 0

$\Sigma I=0$ . By convention, the sign of current entering a junction is positive and current leaving a junction is negative.
Thus, at junction A,

4 + 5 - 6 = 3 A

$4+5-6=3A$ . It is positive so the current 3A will flow from A to B.

The resistance of a copper wire and an iron wire at 20C are 4.

Ω

$\Omega$ and 3.9

Ω

$\Omega$ respectively. Neglecting any thermal expansion, find the temperature at which resistances of bolh are equal.

α_{c u} = 4.0 \times 10^{- 3} K^{- 1}

$\alpha _{ cu }=4.0\times { 10 }^{ -3 }{ K }^{ -1 }$ . and

α_{f e} = 4.0 \times 10^{- 3} K^{- 1}

$\alpha _{ fe }=4.0\times { 10 }^{ -3 }{ K }^{ -1 }$ .

0%
${ 65 }^{ o }C$
0%
${ 75 }^{ o }C$
0%
${ 270 }^{ o }C$
0%
${ 95 }^{ o }C$

Name the quantity which is measured in

e V

$eV$ .

0%
Power
0%
Energy
0%
Charge
0%
Electric potential

Sensitivity of potentiometer can be increased by

0%
increasing the e.m.f. of the cell.
0%
increasing the length of the potentiometer.
0%
decreasing the length of the potentiometer wire.
0%
none of these.

The current

I

$I$ and voltage

V

$V$ graphs for a given metallic wire at two different temperatures

T_{1}

$T_1$ and

T_{2}

$T_2$ are shown in the figure. It is concluded that :

0%
$T_1 > T_2$
0%
$T_1 < T_2$
0%
$T_1 = T_2$
0%
$T_1 = 2 T_2$

Explanation

Slope of the graph will give us reciprocal of resistance. Here resistance at temperature

T_{1}

$T_1$ is greater than that of

R_{2}

$R_2$ . Since resistance of metallic wire is more at higher temperature than at lower temperature, hence

T_{1} > T_{2}

$T_1 > T_2$

If in the experiment of Wheatstone's bridge, the positions of cells and galvanometer are interchanged, then balance point will

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

Explanation

For balanced Wheatstone bridge which is shown in figure a,

\frac{P}{Q} = \frac{S}{R}

$\dfrac{P}{Q}=\dfrac{S}{R}$
If we interchange the cell and galvanometer then circuit becomes as shown in figure b.
and balanced condition,

\frac{P}{S} = \frac{Q}{R} \Rightarrow \frac{P}{Q} = \frac{S}{R}

$\dfrac{P}{S}=\dfrac{Q}{R} \Rightarrow \dfrac{P}{Q}=\dfrac{S}{R}$
Thus, balanced point remains unchanged.

A potentiometer consists of a wire of length 4m and resistance 10

Ω

$\Omega$ . It is connected to a cell of e.m.f. 3V. The potential gradient of wire is

0%
0.75 V/m
0%
2 V/m
0%
6 V/m
0%
10 V/m

Explanation

Step 1: Potentiometer Figure

$\textbf{Step 1: Potentiometer Figure}$

Using KVL in above simple circuit of 1 loop, we get current

I = \frac{V}{R} = \frac{3}{10} = 0.3 A

$I = \dfrac{V}{R} = \dfrac{3}{10} = 0.3\ A$

AB is potentiometer wire

From figure we can say that potential difference across AB will be equal to e.m.f of cell.

V_{A B} = E = 3 V

$V_{AB} = E = 3\ V$ , Also

V_{A B} = I R

$V_{AB} = IR$

\Rightarrow V_{A B} = 0.3 \times 10 = 3 V

$\Rightarrow\ \ V_{AB} = 0.3 \times 10 = 3\ V$

Step 2: Potential gradient calculation

$\textbf{Step 2: Potential gradient calculation}$

Potential gradient

K = \frac{V_{A B}}{L}

$K = \dfrac{V_{AB}}{L}$

= \frac{3}{4} = 0.75 V / m

$= \dfrac{3}{4} = 0.75\ V/m$

Hence, Option

A

$A$ is correct.

2110541_203120_ans_6baf565b22db48ecb5a285308402c9c2.png

Substances which allow electricity to pass through them are known as

0%
Electrical conductors
0%
Electrical insulators
0%
Both conductors & insulators
0%
None of these

State whether true or false.

Bulbs C and D are connected in series.

0%
True
0%
False

Substances that do not allow free flow of charge through them are called __________.

0%
Insulators
0%
Conductors
0%
Both A and B
0%
None of these

The principle involved in potentiometer is

0%
variation of current with variati wire on in the diameter of the potentiometer
0%
similar to the principle of Wheatstone bridge
0%
variation of resistance with temperature
0%
both (a) and (b)

The given figure is a 'material tester'.
Which of the following objects will make the bulb glow when put on the gap shown?

If we touch a naked current carrying wire, we get a shock. This is because our body is a :

0%
conductor of electricity
0%
insulator of electricity
0%
source of electricity
0%
both B&C

Value of Current i in the following circuit is :-

0%
13 A
0%
12 A
0%
9 A
0%
None of these

Explanation

According to Kirchhoff's first law, total entering current into the circuit

=

$=$ total leaving current from the circuit. i.e,

10 + 1 + 2 = i

$10+1+2=i$ or

i = 13 A

$i=13 A$

A group of N cells whose emf varies directly with the internal resistance as per the equation

E_{N} = 1.5 r_{N}

$E_{N}=1.5r_{N}$ are connected as shown in the figure. The current I in the circuit is :-

0%
5.1 A
0%
0.51 A
0%
1.5 A
0%
0.15 A

Explanation

If the current I flows in anticlockwise direction,

I = \frac{E_{1} + E_{2} + . . . . . . . + E_{N}}{r_{1} + r_{2} + . . . . . . + r_{N}} = \frac{1.5 r_{1} + 1.5 r_{2} + . . . . . + 1.5 r_{N}}{r_{1} + r_{2} + . . . . . . + r_{N}} = 1.5 A

$I=\frac{E_1+E_2+.......+E_N}{r_1+r_2+......+r_N}=\frac{1.5r_1+1.5r_2+.....+1.5r_N}{r_1+r_2+......+r_N}=1.5 A$

In a typical Wheatstone network the resistance in cyclic order are

A = 10 Ω, B = 5 Ω, C = 4 Ω

$A=10\:\Omega ,\:B=5\:\Omega ,\:C=4\: \Omega$ and

D = 4 Ω

$D=4\:\Omega$ for the bridge to be balanced :-

0%
$10\:\Omega$ should be connected in parallel $A$
0%
$10\:\Omega$ should be connected in series with $A$
0%
$5\:\Omega$ should be connected in series with $B$
0%
$5\:\Omega$ should be connected in parallel with $B$

Explanation

a) here,

A = 10 | | 10 = 10 / 2 = 5, B = 5, C = 4, D = 4

$A=10||10=10/2=5, B=5,C=4,D=4$

and

A / B = 1, C / D = 1

$A/B=1, C/D=1$ so,

A / B = C / D

$A/B=C/D$

b) here,

A = 10 + 10 = 20, B = 5, C = 4, D = 4

$A=10+10=20, B=5,C=4,D=4$

and

A / B = 4, C / D = 1

$A/B=4,C/D=1$ so

, A / B \neq C / D

$, A/B \neq C/D$

c) here,

A = 10, B = 5 + 5 = 10, C = 4, D = 4

$A=10,B=5+5=10,C=4,D=4$

and

A / B = 1, C / D = 1

$A/B=1,C/D=1$ so,

A / B = C / D

$A/B=C/D$

d)here,

A = 10, B = 5 | | 5 = 5 / 2 = 2.5, C = 4, D = 4

$A=10,B=5||5=5/2=2.5,C=4,D=4$

And

A / B = 4, C / D = 1

$A/B=4,C/D=1$ so,

A / B \neq C / D

$A/B \neq C/D$

At what temperature will the resistance of a copper wire become three times its value at

0^{\circ} C

$0^{\circ}C$ [Temperature coefficient of resistance for copper

= 4 \times 10^{- 3} p e r^{\circ} C

$=4\times 10^{-3}per^{\circ}C$ ] :-

0%
$400^{\circ}C$
0%
$450^{\circ}C$
0%
$500^{\circ}C$
0%
$550^{\circ}C$

Explanation

As we know,

R = R_{0} (1 + α Δ T)

$R=R_0(1+\alpha \Delta T)$

give,

R = 3 R_{0}

$R=3R_0$ and

α = 4 \times 10^{- 3} p e r^{0} C

$\alpha =4\times 10^{-3}per^0C$

putting Values,

3 R_{0} = R_{0} (1 + 4 \times 10^{- 3} (T_{2} - T_{1}))

$3R_0=R_0(1+4\times 10^{-3}(T_2-T_1))$

2 = 4 \times 10^{- 3} (T_{2} - 0)

$2=4\times 10^{-3}(T_2-0)$ as

T_{1} = 0^{0} C

$T_1=0^0C$

therefore,

T_{2} = 500^{0} C

$T_2=500^0C$

Referring to the circuit,

R

$R$ is the resistance of a potentiometer. As the sliding contact is moved from

a

$a$ to

b

$b$ the reading in the ideal ammeter will :

0%
increase
0%
decrease
0%
initial decrease and then increase
0%
initially increase and then decrease

Explanation

R_{a c}

$R_{ac}$ increases as x increases therefore current through ammeter decreases.

R_{a c} =

$R_{ac}=$ Resistance from point a to c

1171696_290232_ans_0ae746fac93d43ac9f79f8fdd3ba7c1c.JPG

Which of the statement is wrong :-

0%
when all resistance are equal, then the sensitivity of wheatstone bridge is maximum.
0%
when the galvanometer and the cell are interchanged, then the balancing of wheat stone bridge will be effected.
0%
Kirchoff's first law for the currents meeting at the Junctions in an electric circuit shows the conservation of charge.
0%
Rheostat can be used as potential divider.

Explanation

Before interchanging cell and galvanometer (fig-a): the null condition is

\frac{P}{Q} = \frac{S}{R}

$\frac{P}{Q}=\frac{S}{R}$
After interchanging cell and galvanometer (fig-b): the null condition is

\frac{P}{S} = \frac{Q}{R} \Rightarrow \frac{P}{Q} = \frac{S}{R}

$\frac{P}{S}=\frac{Q}{R} \Rightarrow \frac{P}{Q}=\frac{S}{R}$
Thus, in both case balanced condition is same so it will not be effected after interchanging cell and galvanometer.

For the following circuit the potential difference between x and y in volt is :-

0%
10
0%
50
0%
100
0%
0

Explanation

Apply Kirchhoff's law
for loop pxqp:

I_{x} (100 + 100) = 200 \Rightarrow I_{x} = 1 A

$I_x(100+100)=200 \Rightarrow I_x=1 A$
for loop pyqp:

I_{y} (100 + 100) = 200 \Rightarrow I_{y} = 1 A

$I_y(100+100)=200 \Rightarrow I_y=1 A$
thus,

V_{p} - V_{x} = 100 I_{x} = 100 V

$V_p-V_x=100I_x=100 V$ and

V_{p} - V_{y} = 100 I_{y} = 100 V

$V_p-V_y=100I_y=100 V$
so,

V_{x} - V_{y} = 0

$V_x-V_y=0$

Electromotive force of a cell is basically a

0%
force
0%
power
0%
work
0%
current capacity

Explanation

EMF

=

$=$ work done per unit charge

= \frac{F d}{q} = \frac{q E d}{q} = E d = V

$= \dfrac{Fd}{q}=\dfrac{qEd}{q}=Ed=V$

How many electrons flow per sec. through the filament of a 220 V and 110 W electric bulb :-

0%
$3.125\times 10^{18}$
0%
$3.125\times 10^{17}$
0%
$312.5\times 10^{18}$
0%
$31.25\times 10^{18}$

Explanation

If $n$ be the number of electrons , then
number of electron per sec $=ne =$ current (I).
now , $P=VI , I=P/V$
thus, $n=I/e=\dfrac{P}{eV}=\dfrac{110}{1.6\times 10^{-19}\times 220}=3.125\times 10^{18}$

Figure

(a)

$(a)$ below shows a Wheatstone bridge in which

P, Q, R, S

$P, Q, R, S$ are fixed resistances,

G

$G$ is a galvanometer and

B

$B$ is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positions of

B

$B$ and

G

$G$ are interchanged,. as shown in figure

(b)

$(b)$ . The new deflection of the galvanometer.

0%
Is to the left.
0%
Is to the right.
0%
Is zero.
0%
Depends on the values of $P, Q, R, S$ .

R_{1}

$R_1$ and

R_{2}

$R_2$ are the filament resistances of a

200 W

$200 W$ bulb and a

100 W

$100 W$ bulb respectively designed to operate on the same voltage, then:

0%
$R_{1}= 2R_{2}$
0%
$R_{2}= 2 R_{1}$
0%
$R_{2}= 4R_{1}$
0%
$R_{1}=4R_{2}$

Explanation

The answer is B.

The power P = VI, substituitng I=V/R we get,

P = \frac{V^{2}}{R}

$P=\dfrac { { V }^{ 2 } }{ R }$ .
Here, the power P1 used in case of R1 is

P 1 = \frac{V^{2}}{R 1} = 200 W

$P1=\dfrac { { V }^{ 2 } }{ R1 } =200W$ .
The power P2 used in case of R2 is

P 2 = \frac{V^{2}}{R 2} = 100 W

$P2=\dfrac { { V }^{ 2 } }{ R2 } =100W$ .

\frac{P 1}{P 2} = \frac{R 1}{R 2} = \frac{1}{2}

$\dfrac { P1 }{ P2 } =\dfrac { R1 }{ R2 } =\dfrac { 1 }{ 2 }$ .
Hence, R2 = 2R1.

Which one of the following will not conduct electricity?

0%
solid NaCl
0%
$CuSO_4$ solution
0%
graphite
0%
acidified water

In these three circuits all the batteries are identical and have negligible internal resistance, and all the light bulbs are identical. Rank all 5 light bulbs (A, B, C, D, E) in order of brightness from brightest to dimmest.

0%
A = B = C > D = E
0%
A > B = C > D = E
0%
A = D = E > B = C
0%
A = D = E > B > C
0%
D = E > A > B = C

In the electrolysis of NaCl

0%
$Cl^-$ is oxidised at anode
0%
$Cl^-$ is reduced at anode
0%
$Cl^-$ is reduced at cathode
0%
$Cl^-$ is neither reduced nor oxidised

Explanation

Answer is A.
Electrolysis of an aqueous solution of table salt (NaCl, or sodium chloride) produces aqueous sodium hydroxide and chlorine, although usually only in minute amounts. NaCl (aq) can be reliably electrolysed to produce hydrogen. Hydrogen gas will be seen to bubble up at the cathode, and chlorine gas will bubble at the anode.
As the electricity from the battery passes through and between the electrodes, the water splits into hydrogen and chlorine gas, which collect as very tiny bubbles around the electrode tips. Hydrogen collects around the cathode and chlorine gas collects around the anode.
Hence, in the electrolysis of NaCl,

C l^{-}

$Cl^-$ is oxidized at the anode.

In the following circuit, each resistor has a resistance of

15 Ω

$15 \Omega$ and the battery has an e.m.f. of

12 V

$12 V$ with negligible internal resistance.
When a resistor of resistance

R

$R$ is connected between

D & F

$D \& F$ , no current flows through the galvanometer (not shown in the figure) connected between

C & F

$C \& F$ . Calculate the value of

R

$R$ .

0%
$10 \Omega$
0%
$15 \Omega$
0%
$5 \Omega$
0%
$30 \Omega$

Explanation

Voltage at point C

\Rightarrow \frac{V_{c} - 0}{15} = \frac{12 - V_{c}}{30}

$\Rightarrow \frac {V_{c}-0}{15}=\frac {12-{V_{c}}}{30}$

2 V_{c} = 12 - V_{c}

$2V_c=12-V_c$

3 V_{c} = 12

$3V_c=12$

V_{c} = 4 v o l t

$V_c=4volt$
It is given that there is no any current through line CF

\Rightarrow V_{c} = V_{F}

$\Rightarrow V_c=V_F$

\Rightarrow V_{F} = 4 v o l t

$\Rightarrow V_F=4volt$

\Rightarrow I_{2} = \frac{12 - V_{F}}{15} = \frac{12 - 4}{15} = \frac{8}{15} A

$\Rightarrow I_2= \dfrac {12-{V_{F}}}{15}=\dfrac {12-4}{15}= \dfrac {8}{15}A$

Now ohm's law for unknown resistance

R_{1} \Rightarrow Δ V = I R_{1}

$R_1 \Rightarrow {\Delta}V=IR_1$

(4 - 0) = (\frac{30}{30 + R_{1}}) (\frac{8}{15}) R_{1}

$(4-0)=(\dfrac {30}{30+R_1})(\dfrac {8}{15})R_1$

4 = \frac{16 R_{1}}{30 + R_{1}}

$4=\dfrac {16R_1}{30+R_1}$

120 + 4 R_{1} = 16 R_{1}

$120+4R_1=16R_1$

12 R_{1} = 120

$12R_1=120$

R_{1} = 10 Ω

$R_1=10{\Omega}$

Two long straight cylindrical conductors with resistivities

ρ_{1}

$\rho_1$ and

ρ_{2}

$\rho_ 2$ respectively are joined together as shown in figure. If current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is

0%
zero
0%
$\displaystyle \frac{(\rho_1 - \rho_2) I \varepsilon_0}{2}$
0%
$\varepsilon_0 I |\rho_1 - \rho|$
0%
$\varepsilon_0 I |\rho_1 + \rho|$

Explanation

Apply Gauss's law :

\frac{q_{i n}}{ε_{0}} =

$\displaystyle \frac{q_{in}}{\varepsilon_0} =$ (outgoing flux - incoming flux)

= Δ (E A)

$= \Delta (EA)$

V = I R = \frac{I ρ l}{A} \Rightarrow \frac{V}{l} A = I ρ \Rightarrow E A = I ρ

$\displaystyle V = IR = \frac{I \rho l}{A} \Rightarrow \frac{V}{l} A = I \rho \Rightarrow EA = I \rho$

\Rightarrow \frac{q_{i n}}{ε_{0}} = I | ρ_{1} - ρ_{2} | \Rightarrow q_{i n} = I ε_{0} | ρ_{1} - ρ_{2} |

$\Rightarrow \displaystyle \frac{q_{in}}{\varepsilon_0} = I |\rho_1 - \rho_2| \Rightarrow q_{in} = I \varepsilon_0 |\rho_1 - \rho_2|$

An electric iron of heating element of resistance

88 Ω

$88 \Omega$ is used at

220

$220$ volt for

2

$2$ hours. The electric energy spent, in unit, will be:

0%
$0.8$
0%
$1.1$
0%
$2.2$
0%
$8.8$

Explanation

Answer is B.

The electric power required to operate a device is the input voltage times the current required.
P = VI
where
P = electric power
V = voltage used
I = current in amperes
VI is V times I
Electric power is measured in watts. The electrical energy used is the electric power times the time. That is, E = Pt. If we measure the electric power as kilowatts and the time as hours, we get the energy used by an electrical system in terms of kilowatt-hours. That is the unit of measurement the electric company uses when determining our bill.
In this case, th electric iron of resistance 88 is used at 220 volt for 2 hours.
The power P = VI, substituitng I=V/R we get,

P = \frac{V^{2}}{R} = \frac{220^{2}}{88} = 550 W = 0.55 k W

$P=\frac { { V }^{ 2 } }{ R } =\frac { { 220 }^{ 2 } }{ 88 } =550W\quad =\quad 0.55kW$ .
Therefore, energy spent = E = Pt =

0.55 k W \times 2 h r s = 1.1 u n i t s

$0.55kW\times 2hrs=1.1\quad units$ .
Hence, the energy spent is 1.1 units.

Choose the correct option

0%
rubber rod becomes positive and fur becomes negative
0%
both rod becomes positive
0%
rubber rod becomes negative and fur becomes positive
0%
both becomes neutral

In a battery five dry cells each of 1.5 volt have internal resistance of

0.2, 0.3, 0.4, 0.5

$0.2, 0.3, 0.4, 0.5$ and

0.6 Ω

$0.6\Omega$ are present in series. The battery is connected to a

1 Ω

$1 \Omega$ resistance. Identify the correct statement (s) :

0%
Current in the circuit will be $2.5 A$
0%
Current in the circuit will be $1.5 A$
0%
On short circuiting the battery $3.75A$ current will flow
0%
Both $(A)$ and $(C)$

Explanation

R_{e q} = r_{1} + r_{2} + r_{3} + r_{4} + r_{5} + R = 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 1

$R_{eq}=r_1+r_2+r_3+r_4+r_5+R=0.2+0.3+0.4+0.5+0.6+1$

R_{e q} = 3 Ω

$R_{eq}=3\Omega$

E_{e q} = 5 E = 7.5 V

$E_{eq}=5E=7.5V$

I = \frac{E_{e q}}{R_{e q}} = \frac{7.5}{3}

$I=\dfrac{E_{eq}}{R_{eq}}=\dfrac{7.5}{3}$

⟹ I = 2.5 A

$\implies I=2.5A$

When battery is shorted , no current flows through

R

$R$ .

Equivalent resistance becomes

R_{e q} = r_{1} + r_{2} + r_{3} + r_{4} + r_{5}

$R_{eq}=r_1+r_2+r_3+r_4+r_5$

⟹ R_{e q} = 3 Ω

$\implies R_{eq}=3\Omega$

Current=

I = \frac{E_{e q}}{R_{e q}}

$I=\dfrac{E_{eq}}{R_{eq}}$

⟹ = I = \frac{7}{2} = 3.5 A

$\implies=I=\dfrac{7}{2}=3.5A$

Answer-(D)

The diagram shows a resistor connected to a cell of e.m.f.

2 V

$2 V$ How much heat energy is produced in the resistor in six seconds?

0%
$2.5 J$
0%
$4.8 J$
0%
$10 J$
0%
$60 J$

Explanation

Answer is B.

The energy dissipated is calculated from the formula E = Pt. That is, E = VI*t.
Substituting I=V/R, we get,

E = \frac{V^{2}}{R} \times t

$E=\dfrac { { V }^{ 2 } }{ R } \times t$ .
In this case, the resistance is 5 ohms and voltage is 2 V.
So,

$E=\dfrac { { 2 }^{ 2 } }{ 5 } \times 6=4.8\quad J$ .
Hence, the heat energy is produced in the resistor in six seconds is 4.8 J.

Two bulbs, one of 200 W and the other of 100 W, are connected in series with a 100 V battery which has no internal resistance. Then,

0%
The current passing through the 200W bulb is more than that through the 100W bulb
0%
The power dissipation in the 200W bulb is more than that In the 100 W bulb
0%
The voltage drop across the 200W bulb is more than that across the 100W bulb
0%
The power dissipation In the 100W bulb is more than that in the 200W bulb

On a bulb is written 220 Volt and 60 watt. Find out the resistance of the bulb and the value of the current flowing through it.

0%
806.66 ohm 0.27 ampere
0%
500 ohm 12 ampere
0%
200 ohm 14 ampere
0%
100 ohm 11 ampere

Explanation

$P=\dfrac{V^2}{R}=VI$

Hence,

$R=\dfrac{V^2}{P}=\dfrac{220\times 220}{60}$

$\implies R=806.66\Omega$

$I=\dfrac{P}{V}=\dfrac{60}{220}$

$\implies I=0.27A$

Answer-(A)

How many electrons constitute current of one micro ampere in one second?

0%
$6.25\times10^8$
0%
$6.25\times10^{12}$
0%
$6.25\times10^9$
0%
$6.25\times10^{15}$

Explanation

Given: Current ,

$I = 1\mu A = 10^{-6}A$
We know that

$Q = ne, t = 1s$
Where

$n$ is the number of electrons and e is the charge on

$1$ electron and

$e = 1.6\times10^{-19}C$

$n = \displaystyle\frac{I}{e} = \displaystyle\frac{10^{-6}}{1.6\times10^{-19}} = 6.25\times10^{12} Electrons$

what is

$i$ ?

0%
$-4$ mA
0%
$+2$ mA
0%
$+4$ mA
0%
$+8$ mA

Figure below shows a portion of an electric circuit with the currents in ampere and their directions. The magnitude and direction of the current in the portion

$PQ$ is:

0%
$0A$
0%
$3A$ from $P$ to $Q$
0%
$4A$ from $Q$ to $P$
0%
$6A$ from $Q$ to $P$

Explanation

Using Kirchoff's current law, which states that the net current in a junction is zero
the let The magnitude of the current in the portion $$PQ$$ is

$x$ and the direction P to Q.

$x=-3A+1A-2A-8A+4A+2A$

$x=-13A+7A= - 6A$
hence,the magnitude of the current in the portion

$PQ$ is

$6A$ and the direction from

$Q$ to

$P$ .

In which of the following cells, the potential difference between the terminals of a cell exceeds its emf.

0%
$a$
0%
$b$
0%
$c$
0%
$d$

An electric iron draws a current of 15 A from a 220 V supply, What is the cost of using iron for 30 min everyday for 15 days if the cost of unit (1 unit =1 kWhr) is 2 rupees ?

0%
Rs 49.5
0%
Rs 60
0%
Rs 40
0%
Rs 10

Explanation

As we know that power(P)=v

$\times$ I
so P=15

$\times$ 220=3300 watt
cost per unit=2 rupees
therefore,

$\dfrac {3300\times2\times15\times30}{60\times1000}$ =49.5 rupees.

The circuit given below is for the operation of an industrial fan. The resistance of the fan is

$3 ohm$ . The regulator provided with the fan is a fixed resistor and a variable resistor in parallel. Under what value of the variable resistance given, power transferred to the fans will be maximum? The power source of the fan is a dc source with internal resistance of

$6 ohm$

0%
3 $\Omega$
0%
$0$
0%
$\infty$
0%
$6$ $\Omega$

Explanation

The power transferred to fan is given as

$P=V^2/R$ , where

$R$ is the total resistance of the circuit. As power is inversely proportional to total resistance. So for maximum power, the total resistance should be minimum. Total resistance here is

$R=\dfrac{6r}{6+r}+3$ .

$r$ is the variable resistance.

$R$ is minimum when

$r=0$ .

The number of joules contained in

$1 kWh$ is

0%
$36 \times {10}^{2}$
0%
$36 \times {10}^{3}$
0%
$36 \times {10}^{4}$
0%
$3.6 \times {10}^{6}$

Explanation

1 KWh = 1000

$\ast$ joule/sec

$\ast$ 60

$\ast$ 60 sec

1KWh = 1000

$\ast$ 60

$\ast$ 60joue/sec

$\ast$ sec

1KWh = 1000

$\ast$ 60

$\ast$ 60 joule

1KWh = 36,00,000 joule = 3.6

$\ast$ 10

$^{6}$

An electric heater has a rating of 2 kW, 220 V. The cost of running the heater for 10 hours at the rate of Rs 350 per unit is Rs ____

0%
216
0%
165
0%
209
0%
105

Explanation

Power is energy consumed per unit time.

Here, energy consumed

$=P=2kW=2000W$ and time

$=t=10hr=36000s$

$P=\dfrac{E}{t}$

$\implies E=Pt=2kW\times 10h=20kWh=20units$

Since

$1unit=1kWh$

Cost

$=20\times 350=Rs 7000$

In the circuits shown below the ammeter

$A$ reads

$4 amp$ and the voltmeter

$V$ reads

$20 volts$ . The value of the resistance

$R$ is

0%
Slightly more than 5 ohms
0%
Slightly less than 5 ohms
0%
Exactly 5 ohms
0%
None of the above

Explanation

$i=i_{1}+i_{2}$ (Kirchoff Juction Law)

$i_{1}R=V$ (ohm law)

Given

$V=20$

$i=4A$

$(i-i_{2})R=20$

$R=(\dfrac{20}{4-i_{2}})\Omega$

The value of R is slightly more than

$5\Omega.$

A source of electromotive force (emf) is a:

0%
Cell
0%
Battery
0%
Generator
0%
All of these

Calculate the resistance offered by 3 HP water pump which runs on 220 V supply

0%
21.63 $\Omega$
0%
42.6 $\Omega$
0%
63.2 $\Omega$
0%
37.6 $\Omega$

Explanation

Power

$=P=3HP=3\times 746W=2238W$

And,

$p=\dfrac{V^2}{R}$

$\implies R=\dfrac{V^2}{P}=\dfrac{220\times 220}{2238}$

$\implies R=21.63\Omega$

Answer-(A)

Figure below shows a portion of an electric circuit with the currents in ampreres and their directions. The magnitude and direction of the current in the portion PQ is :

0%
0A
0%
3A from P to Q
0%
4A from Q to P
0%
6A from Q to P

CBSE Questions for Class 12 Medical Physics Current Electricity Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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