1. (d) 25

2. (b) IR2

3. (d) 25 W

4. (c)1:4

5. A
voltmeter is always connected in parallel to resistance across the point
between

which the
potential difference is to be measured.

6. Diameter
of wire (d) = 0.5 mm, resistivity (ρ) 1.6 x 10-8 Ωm, resistance (R) = 10 Ω.

R =
ρL/A

L=
πD2R/4ρ

= 22 x
(5 x 10-4)

2/ 7 x 4 x
1.6 x 10-8 = 122.5 m

If the
diameter is doubled for given length of given material resistance is inversely

proportional
to the cross-section area of wire.

7. From the
given data the I-V graph is a straight line

Resistance
of resistor (R) = VA-VB/1A-1B = 12 V – 6 V/ 3.6 A – 1.8 A

= 6V/
1.8 A = 3.3 Ω

8. Voltage
of battery = V = 12 V, Current ( I) = 2.5 mA = 2.5 x 10-3 A

Resistance
(R) = V/I = 12V/ 2.5 x 10-3 A = 4800 Ω.

9. Potential
difference (V) = 9 V.

Total
resistance (R) = R1+ R2+ R3+R4 +R5

= 0.2
+0.3 + 0.5 + 0.5 + 12 = 13.4 Ω

Current
in the circuit (I) = V/R = 9 V / 13. 4 Ω = 0.67 A.

In series
circuit same current flows through all the resistance, hence current of 0.67

A will flow
through 12 Ω resistor.

10. Let a
resistors of 176Ω are joined in parallel. Then their combined resistance (R)

1/R =
1/176 + 1/176 …… times = n/176 or R = 176/n Ω

It is
given that V= 220 V and I = 5 A

R =
V/I or 176/n = 220/5 = 44Ω

n = 176/44 =
4, 4 resistors should e joined in parallel.

12. Each
bulb is rated as 10 W, 220 V, It draws a current (I) = P/V = 10W/220 V

= 1/22
A.

As the
maximum allowable current is 5A and all lamps are connected in parallel,

hence
maximum number of bulbs joined in parallel with each other = 5 x 22 = 110.

13. It is
given that potential difference (V) = 220 V.

Resistance
of coil (A) = Resistance of coil (B) = 24 Ω

(i) When
either coil is used separately, the circuit (I) = V/R = 220 V/ 24 Ω

= 9.2 A.

(ii) When
two coils are used in series total resistance (R)

= R1 +
R2 = 24 +24 = 48Ω

Current
flowing (I) = V/ R = 220 V/ 48 Ω = 4.6 A.

(iii) When
two coils are joined in parallel. Total resistance (R) = 1/24 + 1/24

= 2/24, R =
12Ω.

Current (I)
= V/R = 220V / 12Ω = 18.3 A.

14. (i) When
a 2Ω resistor is joined t a 6 V battery in series with 1Ω and 2Ω resistors.

Total
resistance (R) = 2 + 1 + 2 = 5Ω.

Current (I)
= 6V/5Ω = 1.2 A

Power used
in 2 A resistor = I2R = 2.88 W

(ii) When 2Ω
resistor is joined to a 4 V battery in parallel with 12Ω resistor and 2Ω

resistors,
the current flowing in 2Ω = 4 V/ 2Ω = 2 A/.

Power used
in 2Ω resistor = I2R = 8 W

Ratio =
2.88/8 = 0.36: 1.

15. Current
drawn by 1st lamp rated 100 W at 220 V = P/V = 100/ 220 = 5/11 A.

Current
drawn by 2nd lamp rated 60 W at 220 V = 60/220 = 3/11 A.

In
parallel arrangement the total current = I1 +I2 = 3/11+ 5/11 = 8/11 = 0.73 A.

16. Energy
used by a TV set of power 250 W in 1 hour = P x t = 250 Wh.

Energy
used by toaster of power 1200 W in 10 minute (10/60 h)

= P x
t = 1200 W x 10/60 h = 200 Wh. Material downloaded from http://myCBSEguide.com
and http://onlineteachers.co.in

Portal for
CBSE Notes, Test Papers, Sample Papers, Tips and Tricks

17.
Resistance of electric heater (R) = 8Ω, current (I) = 15 A.

Rate
at which heat developed in the heater = I2Rt/t = 15 x 15 x 8 = 1800 W.

18. (a) For
filament of electric lamp we require a strong metal with high melting

point.
Tungsten is used exclusively for filament of electric lamps because its

melting
point is extremely high.

(b)
Conductors of electric heating devices are made of an alloy rather than a pure

metal due to
high resistivity than pure metal and high melting point to avoid

getting
oxidized at high temperature.

(c) Series
arrangement is not used for domestic circuits as current to all

appliances
remain same in spite of different resistance and every appliance

can not be
switched on/ off independently.

(d)
Resistance of a wire is inversely proportional to its cross-section area.

(e) Copper
and aluminium wires are usually employed for electricity

transmission
because they are good conductor with low resistivity. They are

ductile also
to be drawn into thin wires.

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