[3 Mark Question Answer]

Question 13 Marks

A bulb marked 12 V, 24 W operates on a 12 V battery for 20 minutes. Calculate:

(i) The current flowing through it, and

(ii) The energy consumed.

Answer

(a) Given,
Power, $P=24 W$
Voltage, $V=12 V$
Current, $I=$ ?
As power, $P = VI$
(i) $I =\frac{24}{12}=2 A$
(ii) Energy, $E = P \times t$
$E=24 \times 20 \times 60 sec$
$E=28800 J$

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Question 23 Marks

An electrical appliance having a resistance of 200 Ω is operated at 220 V. calculate the energy consumed by the appliance in 5 minutes.
in joules.

Answer

Given, Resistance , $R=200 \Omega$
Voltage, $V=200 V$
time, $t=5 min =5 \times 60 sec =300 sec$
As, Energy , $E=\frac{ V ^2 t }{ R }$
In joules
$E=\frac{(200)^2 \times 300}{200}$
$E=60000 J$

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Question 33 Marks

An electric bulb is rated at 220 V, 100 W.
(a) What is its resistance?
(b) What safe current can be passed through it?

Answer

(a) Given,
Power P $=100 W$
Voltage V $=220 V$
As power, $P=\frac{V^2}{R}$
$
R=\frac{(220)^2}{100}=484 \Omega
$
(b) The safe limit of current that can pass through it
$\begin{aligned} & I=\frac{P}{V} \\ & I=\frac{100}{220}=0.45 A \end{aligned}$

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Question 43 Marks

An electric bulb is rated '100 W, 250 V'. How much current will the bulb draw if connected to a 250 V supply ?

Answer

Given,
Power P $=100 W$
Voltage V $=250 V$
As Power , $P = VI$
$
I =\frac{100}{250}=0.4 A
$

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Question 53 Marks

Calculate the current through a 60 W lamp rated 250 V. If the line voltage falls to 200 V, how is power consumed by the bulb affected?

Answer

Given,
Power $P =60 W$
Voltage, $V=250 V$
As Power, $P = VI$
$
I =\frac{60}{250}=0.24 A
$
Resistance of bulb $R =\frac{ V ^2}{ P }=\frac{250^2}{60}=1041.6 \Omega$
Now if voltage falls to $200 V$, power consumed will be
$P=\frac{V^2}{R}=\frac{200^2}{1041.6}=38.4 W$

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Question 63 Marks

A current of 2 A is passed through a coil of resistance $75 \Omega$ for $2$ minutes.
(a) How much heat energy is produced?
(b) How much charge is passed through the resistance?

Answer

Given, current $( I )=2$ AResistance $R =75 \Omega$
Time, $t =2 min=120 s$
(a) Heat produced, $H =1^2 Rt$
or, $H =(2)^2(75)(120) J =36000 J$
(b) Charge passed, $Q = It$
or, $Q=(2)(120) C=240 C$

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Question 73 Marks

An electric kettle is rated 2.5 kW, 250 V. Find the cost of running the kettle for two hours at Rs. 5.40 per unit.

Answer

Given ,

Power of kettle , P = 2.5 kW

Voltage , V = 250 V

Time, t = 2 h

As , Energy , E = P × t

= 2.5 × 2 = 5 kWh

Cost per unit of energy = Rs 5.40

Cost for 5 kWh of energy = 5.40 × 5 = Rs 27

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Question 83 Marks

An electric bulb of resistance 500 Ω draws current 0.4 A from the source. Calculate:
(a) the power of bulb and
(b) the potential difference at its end.

Answer

Resistance of electric bulb (R) = 500Ω

Current drawn from the source (I) = 0.4 A

Power of the bulb (P) = VI

V = I x R

V = 0.4 x 500 = 200 V

The potential difference at its end is 200 V.

Hence,

Power (P) = VI

P = 200 x 0.4 = 80 W

The power of the bulb is 80 Watt.

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Question 93 Marks

An electric toaster draws 8 A current in a 220 V circuit. It is used for 2 hr. Find the cost of operating the toaster if the cost of electrical energy is Rs. 4.50 per kWh.

Answer

Given:
Voltage (V) = 220 V
Current (I) = 8 A
Time (t) = 2 h

Energy (E) = Vlt = 220 × 8 × 2 = 3.52 kWh

>The cost per kWh is Rs. 4.50 >Hence, the cost of running the kettle for two hours is: >3.52 × 4.50 = Rs. 15.84

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Question 103 Marks

Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?

Answer

$
P=\frac{V^2}{R}
$
Heat gained $=\frac{ V ^2}{ R } \times t$
$
\left(\frac{ V _1^2}{ R }\right) \times t _1=\left(\frac{ V _2^2}{ R }\right) \times t _2
$
$
t _2=\left(\frac{ V _1}{ V _2}\right)^2 \times t _1
$
$
=\left(\frac{220}{200}\right)^2 \times 300=363 s =6.05 min
$

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Question 113 Marks

A bulb is connected to a battery of p.d. $4 V$ and internal resistance $2.5 \Omega$. A steady current of $0.5 A$ flows through the circuit. Calculate:The energy dissipated in the bulb in $10$ minutes.

Answer

Given ,Voltage , $V = 4 V$
Resistance of the battery , $RB = 2.5 \Omega$
Current ,$ I = 0.5 A$
Energy dissipated in the bulb in 10 min , $E = I^2Rt$
$E = (0.5)^2 \times 5.5 \times 600 = 825 J$

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Question 123 Marks

A bulb is connected to a battery of p.d. $4 V$ and internal resistance $2.5 \Omega$. A steady current of $0.5 A$ flows through the circuit. Calculate:The resistance of the bulb

Answer

Given, Voltage, $V =4 V$
Resistance of the battery, $RB =2.5 \Omega$
Current, $I =0.5 A$
Total resistance, $R =8 \Omega$
Resistance of the battery, $R _{ B }=2.5 \Omega$
Resistance of the blub, $R _{ b }=8-2.5 \Omega=5.5 \Omega$

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Question 133 Marks

A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 Ω . A steady current of 0.5 A flows through the circuit. Calculate:

The total energy supplied by the battery in 10 minutes

Answer

Given,
Voltage, $V =4 V$
Resistance of the battery, $R_B=2.5 \Omega$
Current, $I =0.5 A$
Energy supplied by the battery,$E=\frac{V^2}{R}$
$\begin{aligned} & t =10 \times 60=600 sec \\ & R =\frac{ V }{ I }=\frac{4}{0.5}=8 \Omega \\ & E =\frac{(4)^2 \times 600}{8}=1200 J \end{aligned}$

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Question 143 Marks

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply of the energy supplied in kWh to the three heaters in 5 hours.

Answer

Given ,

Power , P = 250 W

Voltage , V = 100 V

Time for which energy is supplied , t = 5 h

As , Energy E = P × t

E = 250 × 5 = 1250 Wh

or E = 1.25 kWh

Energy for three heaters = 3 × 1.25 = 3.75 kWh

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Question 153 Marks

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. The resistance of each heater

Answer

Given,
Power , P $=250 W$
Voltage, $V=100 V$
Resistance for each heaters $R =\frac{ V }{ I }$
$
=\frac{100}{2.5}=40 \Omega
$

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Question 163 Marks

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate :

The total current taken from the supply

Answer

Given,
Power, $P =250 W$
Voltage, $V=100 V$
Current through each heater, $I=$ ?
As $P = VI$
$
\begin{aligned}
& \Rightarrow I =\frac{ P }{ V } \\
& =\frac{250}{100}=2.5 A
\end{aligned}
$
$\therefore$ Current taken for the three heaters $=3 \times 2.5=7.5 A$

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Question 173 Marks

A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:

(i) The resistance of the wire, and

(ii) The heat energy produced in 1 minute.

Answer

Given,
|Current, $I =0.2 A$
Potential difference, V $=15 V$
Time,$t=60 sec$
As $V = IR$
(a) $R =\frac{15}{0.2}=75 \Omega$
(b) Heat energy, $H = I ^2 Rt$
$
H=(0.2)^2 \times 75 \times 60
$
or $H =180 J$

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Question 183 Marks

Electrical power P is given by the expression $P = (Q \times V) ÷$ time
(a) What do the symbols $Q$ and $V$ represent?
(b) Express the power $P$ in terms of current and resistance explaining the meanings of symbols used there in.

Answer

(a) Q represents Charge and $V$ represents Voltage.
(b) Electrical Power , $P = I^2R$
where I : Current
and R : Resistance

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Question 193 Marks

Four resistors each of resistance 2Ω are connected in parallel. What is the effective resistance?

Answer

$
\begin{aligned}
& R 1=2 ohm \\
& R 2=2 ohm \\
& R 3=2 ohm
\end{aligned}
$
R4 $=2 ohm$
$
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}
$
$
=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2
$
$
R =0.5 ohm
$

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Question 203 Marks

Two resistors having resistance 4𝛀 and 6𝛀 are connected in parallel. Find their equivalent resistance.

Answer

Let ' $R$ be their equivalent resistance of the $4 \Omega$ and $6 \Omega$ resistors connected in parallel
Then, $\frac{1}{R}=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12} \Omega$
Or, $R=\frac{12}{5}=2.4 \Omega$
Z

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Question 213 Marks

A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.

Answer

(a) ε = 15 V
R = 6 + 3 = 9 ohm
r = 3 ohm
I = ?
I = ε / (R + r)
I = 15 / (9 + 3) = 15/12 = 1.25 A
(b) Current (calculated in (a) part) I = 1.25 A
External Resistance R = 6 + 3 = 9 ohm
V = IR = 1.25 × 9 = 11.25 V

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Question 223 Marks

A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig. 8.41

(a) what would be the reading of the ammeter?
(b) what is the potential difference across the terminals of the cell?

Answer

(a) ε = 1.8 V
Total Resistance = 2 + 4.5 + 0.7 = 7.2 W
I = ?
I = ε / R (total resistance)
I = 1.8 / 7.2 = 0.25 A
(b) Current (calculated in (a) part) I = 0.25 A
Now, total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm
V = IR = 0.25 x 5.2 = 1.3 V

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Question 233 Marks

Three resistors of $6.0 ohm , 2.0 ohm$ and $4.0 ohm$ are joined to ammeter $A$ and a cell of emf $6.0 V$ as shown in following figure. Calculate:
(a) the efective resistance of the circuit.
(b) the reading of ammeter

Answer

(a)
$
R _1=6 W
$
$
R^{\prime}=R_2+R_3=2+4=6 W
$
$R_1$ and $R^{\prime}$ in parallel :
$
\frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R ^{\prime}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}
$
$R =3 ohm$
(b)
$
R =3 ohm
$
$\begin{aligned} & V=6 V \\ & I =?\end{aligned}$
$I=\frac{V}{R}=\frac{6}{3}=2 A$

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Question 243 Marks

A particular resistance wire has a resistance of 3 ohm per meter. Find :

The total resistance of three lengths of this wire each 1.5 m long, in parallel.

Answer

$R =3$ ohm for $1 m$
For $5 m : R =3 \times 5=15 ohm$
But Area $A$ is double i.e. $2 A$ and Resistance is inversely proportional to area so Resistance will be half.
$R =\frac{15}{2}=7.5 ohm$

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Question 253 Marks

Calculate current flowing through each of the resistors A and B in the circuit shown in following figure?

Answer

For resistor $A$ :
$
R =1 ohm
$
$
V=2 V
$
$
=\frac{ V }{ R }=\frac{2}{1}=2 A
$
For resistor B :
$
R =2 ohm
$
$
V =2 V
$
$
I=\frac{V}{R}=\frac{2}{2}=1 A
$

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Question 263 Marks

Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.

Calculate current in each resistor.

Answer

Equivalent resistance of the circuit:
$
\begin{aligned}
& \frac{1}{ R }=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12} \\
& \therefore R =\frac{12}{5}=2.4 \Omega
\end{aligned}
$
Hence, the emf of the cell is
$
V=I R
$
$
\therefore V =0.5 \times 2.4
$
$
\therefore V =1.2 V
$
Therefore, current through each resistor is
$
\begin{aligned}
& I _4=\frac{ V }{ R _4}=\frac{1.2}{4}=0.3 A \\
& I _6=\frac{ V }{ R _6}=\frac{1.2}{6}=0.2 A
\end{aligned}
$

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Question 273 Marks

A resistor of 6Ω is connected in series with another resistor of 4 Ω. A potential difference of 20 V is applied across the combination. Calculate (a) the current in the circuit and (b) the potential difference across the 6Ω resistor.

Answer

$\begin{aligned} & \text { (a) } R_1=6 ohm \\ & R_2=4 ohm \\ & R = R \_1+ R \_2=6+4=10 ohm \\ & V =20 V \\ & I= V / R =20 / 10=2 A \\ & \text { (b) } R =6 W \\ & I=2 A \\ & V=? \\ & V=I R=6 \times 2=12 V \end{aligned}$

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Question 283 Marks

The diagram below in Fig. 8.40 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the
voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.

Answer

(i) Ammeter reading = 0 because of no current
Voltage V = ϵ − Ir
V = 2 − 0 × 1 = 2 volt

(ii) Ammeter reading :
I =ε /(R + r)
I=2 / (4+1) = 2 / 5 = 0.4 amp
Voltage reading :
Voltage V = ϵ - Ir
V=2 - 0.4 x 1 = 2 - 0.4 = 1.6 V

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Question 293 Marks

Five resistors, each 3 Ω, are connected as shown in Fig 8.48. Calculate the resistance (a) between the points P and Q. (b) between the points X and Y.

Answer

(a) $R_1=3+3=6 W$
$
R_2=3 W
$
$R \_1$ and $R \_2$ are connected in parallel
$
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
$
(b) As calculated above $R =2 ohm$
$
R 3=3 ohm
$
$
R 4=3 ohm
$
$
R^{\prime}=R+R 3+R 4=2+3+3=8 ohm
$

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Question 303 Marks

Calculate the equivalent resistance between A and B in the adjacent diagram.

Answer

$
\begin{aligned}
& R _1=3+2=5 ohm \\
& R _2=30 W \\
& R _3=6+4=10 ohm
\end{aligned}
$
$R _1, R _2$ and $R _3$ are connected in parallel
$
\begin{aligned}
& \frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R _2}+\frac{1}{ R _3} \\
& =\frac{1}{5}+\frac{1}{30}+\frac{1}{10}=\frac{10}{30}
\end{aligned}
$
$
R =3 ohm
$

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Question 313 Marks

A wire of uniform thickness with a resistance of 27Ω is cut into three equal pieces and they are joined in parallel. Find the equivalent resistance of the parallel combination.

Answer

Wire cut into three pieces means new resistance $=\frac{27}{3}=9 ohm$
Now three resistance connected in parallel :
$
\begin{aligned}
& \frac{1}{ R _{ P }}=\frac{1}{ R _1}+\frac{1}{ R _2}+\frac{1}{ R _3}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9} \\
& Rp _{ p }=\frac{9}{3}=3 ohm
\end{aligned}
$

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Question 323 Marks

In the circuit shown below in Fid 8.43, calculate the value of x if the equivalent resistance between A and B is 4Ω

Answer

$\begin{aligned} & r_1=4 \text { ohm } \\ & r_2=8 \text { ohm } \\ & r_3=\text { xohm } \\ & r_4=5 \text { ohm } \\ & r =4 \text { ohm } \\ & r_{=}=r_{-} 1+r_{-} 2=4+8=12 \text { ohm } \\ & r_{-}=r_{-} 3+r_{-} 4=(x+5) \text { ohm } \\ & \frac{1}{r}=\frac{1}{r}+\frac{1}{r} \\ & \frac{1}{4}=\frac{1}{12}+\frac{1}{5+x} \\ & \frac{1}{6}=\frac{1}{5+x} \\ & x =1 \text { ohm }\end{aligned}$

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Question 333 Marks

A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.

Answer

Resistance of each set:
$
\begin{aligned}
& 1 r =2+2+2=6 ohm \\
& 2 r =2+2+2=6 ohm \\
& 3 r =2+2+2=6 ohm \\
& 4 r =2+2+2=6 ohm
\end{aligned}
$
Now these resistances are arranged in parallel :
$\begin{aligned} & \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4} \\ & \frac{1}{r}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\end{aligned}$
$r=\frac{6}{4}=1.5 ohm$

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Question 343 Marks

Calculate the equivalent resistance of the following combination of resistors $r 1, r 2$, $r 3$ and $r 4$ if $r_1=r_2=r_3=r_4=2.0 \Omega$ between the points $A$ and $B$ in Fig. 8.42

Answer

$\begin{aligned} & r_1=r_2=r_3=r_4=2.0 ohm \\ & r=r_1+r_2=2+2=4 ohm \\ & \frac{1}{r}=\frac{1}{r_3}+\frac{1}{r_4}=\frac{1}{2}+\frac{1}{2}=1 \\ & r =1 ohm \end{aligned}$
$r=r+r=4+1=5 ohm$

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Question 353 Marks

In parallel combination of resistances:

P.D. is same across each resistance
Total resistance is increased
Current is same in each resistance
All of the above are true

Answer

In parallel combination of resistances, P.D. is same across each resistance.

Hint: In parallel combination, the ends of each resistor are connected to the ends of the same source of potential. Thus, the potential difference across each resistance is same and is equal to the potential difference across the terminals of the source (or battery).

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Question 363 Marks

Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

Answer

When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

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Question 373 Marks

A cell of e.m.f ε and internal resistance r is used to send current to an external resistance R. write expresssions for (a) the total resistance of circuit, (b) the current drawn from the cell. (c) the p.d
across the cell. And (d) voltage drop inside the cell.

Answer

(a) Total resistance $=R+r$
(b) Current drawn from the circuit:
As we know that,
$
\begin{aligned}
& \varepsilon=V+V \\
& =I R+I r \\
& =I(R+r) \\
& I=e /(R+r)
\end{aligned}
$
(c) p.d. across the cell : $\frac{\varepsilon}{(R+r)} \times R$
(d) voltage drop inside the cell: $\frac{\in}{R+r} \times r$

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Question 383 Marks

Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.

Answer

Internal resistance of a cell depends upon the following factors:

(i) The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal
resistance.
(ii) The distance between the electrodes: As the distance between the electrodes increases, the internal resistance of cell also increases. It is also affected by the nature, concentration and temperature of the solution (electrolyte).

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Question 393 Marks

Explain the meaning of the terms e.m.f.., terminal voltage, and internal resistance of a cell.

Answer

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).
Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.
Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell

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Question 403 Marks

Calculate the current flowing through a wire of resistance 5 Ω connected to a battery of potential difference 3 V.

Answer

Potential Difference/Voltage (V) = 3 V
Resistance (R) = 5 ohm
Current (I) = ?
According to Ohm's Law :
V=IR
Then I = V/R
I = 3/5 = 0.6 A

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Question 413 Marks

A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more same or less when the bulb is not glowing?

Answer

Potential Difference/Voltage (V) = 12 V
Current (I) = 2 A
Resistance (R) = ?
According to Ohm's Law :
V=IR
Then R = V/I
R = 12 / 2
R = 6 Ohm
Resistance will be less when the bulb is not glowing.

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Question 423 Marks

Find the potential difference required to pass a current of 0.2 A in a wire of resistance 20Ω

Answer

Current (I) = 0.2 A

Resistance (R) = 20 ohm
Potential Difference (V) = ?
According to Ohm's Law :
V = IR
V = 0.2 × 20 = 4 V

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Question 433 Marks

What length of copper wire of resistivity 1.7 × 10-8 𝛀 m and radius 1 mm is required so that its resistance is 1𝛀?

Answer

Resistance $(R)=1 ohm$
Resistivity $(\rho)=1.7 \times 10-8$ ohm metre
Radius $(r)=1 mm =10-3 m$
Length $(l)=$ ?
$\begin{aligned} & R=\rho \frac{1}{A} \\ & I=\frac{R A}{\rho} \\ & =\frac{R \pi r^2}{\rho} \\ & =\frac{1 \times \pi \times 10^{-6}}{1.7 \times 10^{-8}} \\ & =184.7 m \end{aligned}$

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Question 443 Marks

A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?

Answer

Resistance $(R)=9 ohm$
Length I $=30 cm$
New Length $(l)=30 cm =3 / l =10 cm$
$R=\rho \frac{1}{A}$
New Resistance:
With change in length, there will be change in area of cross-section also in the same order.
$\begin{aligned} & R=\rho \frac{\frac{I}{3}}{3 A} \\ & R=\frac{1}{9} \rho \frac{I}{A} \\ & R=\frac{1}{9} R \\ & R =1 ohm \end{aligned}$

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Question 453 Marks

A wire 3 ohm resistance and 10 cm length is stretched to 30 cm length. Assuming that it has a
uniform cross section, what will be its new resistance?

Answer

Resistance $(R)=3$ ohm
Length $I =10 cm$
New Length $\left( l ^{\prime}\right)=30 cm =3 \times l$
$R =p \frac{1}{A}$
New Resistance:
With stretching length will increase and area of cross-section will decrease in the same order
$R ^{\prime}=p \frac{3}{A / 3}$
Therefore,
$
R ^{\prime}=9 p \frac{1}{A}=9 R
$
$
R^{\prime}=9 \times 3=27 \Omega
$

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Question 463 Marks

State Ohm’s law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.

Answer

It states that electric current flowing through a metallic wire is directly proportional to the potential difference $V$ across its ends provided its temperature remains the same. This is called Ohm's law.
$V=I R$

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Question 473 Marks

Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part.

Answer

In metals, free electrons are the moving charges that result in the conduction of electricity. If 'n' electrons pass through the metallic conductor in time 't', then the total charge that has flown is given by Q (charge) = n x e (charge on an electron).

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Question 483 Marks

State whether the potential is a scalar or vector? What does the positive and negative sign of potential convey?

Answer

Potential is also a scalar quantity. The positive sign of potential conveys that work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity. The negative sign of potential conveys that work is done on the negative test charge by the attractive force.

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Question 493 Marks

Write an expression connecting the resistance and resistivity. State the meaning of symbols used.

Answer

Expression :
$
R=p \frac{I}{A}
$
$p$ - resistivity
$R$ - resistance
I - length of conductor
A - area of cross-csection

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Question 503 Marks

How does the resistance of a wire depend on its length? Give a reason for your answer with reason.

Answer

Resistance of a wire is directly proportional to the length of the wire.
R ∝ I
The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.

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