Content from CBSE Science text book for Std 10 , includes numerical problems

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1752, Benjamin Franklin

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7. * P.D SPLITS
* CURRENT REMAINS SAME
* EFFECTIVE RESISTANCE IS
THE SUM OF THE INDIVIDUAL
RESISTANCES OF RESISTORS
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8. * Current splits
* P.D remains same
* Reciprocal of effective
resistance is the sum
of reciprocals of all
the individual
resistances of
resistors
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10.  file:///C:/Users/Sri/Desktop/ohms-
law%20animated.html
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SYSTEM CURRENT Pot. Diff. RESISTANCE
S I Ampere Volt Ohm
1 A = 1000mA
1mA = (1/1000 )A

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Heating effect of
Electric Current
Cell-P.D-Current-Resistance-Heat
1
2
3
4

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13.  The law implies that heat (H) produced in a resistor is
 directly proportional to the square of i for a given R
H α i2
 directly proportional to resistance for a given current
H α R
 and is directly proportional to the passage of time
H α t
By combining all the three expression, we get
H = i2 Rt
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14. 1
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15.  Filment :
-thermally isolated
- max. heat & min.
light
- high melting point
- Tungsten (3380˚c)
 Inert gases
- prolongs the life
time
- Nitrogen, Argon
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16.  protects the circuit
 encased in
porcelain cotridge
 connected in series
 alloy of Tin & Lead
 high resistance
 low melting point
 For domestic needs
1A,2A.3A,5A,10A
fuses
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17. Electrical energy is the
product of power and time.
Its SI unit is watt hour(W h).
One watt hour is the energy
consumed when1 watt of
power is used for 1 hour.
The commercial unit of
Electric Energy is kWh,
1 kW h = 1000 W × 3600 s
= 3.6 × 106 Ws
= 3.6 × 106 joules
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18. Given,
P = 880 W
V = 220 V
We have ,
P = V * I
i = P / V = 880 / 220
I = 4A
As the iron box
consumes 4A, fuse
of 5A rated is to be
used.
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19. Pg : 202
Given,
Charge = Q = 1C
P.D = V = 6V
Energy given = work done = ?
We have ,
W = V * Q
= 6 x 1
= 6 joules.
How much
energy is
given to
each
coulomb of
charge
passing
through a
6 V
battery?
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20. Pg : 213
Draw a
schematic
diagram of a
circuit consisting
of a battery of
three cells of 2 V
each, a 5 W
resistor, an 8 W
resistor, and a
12 W resistor,
and
a plug key, all
connected in
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21. Pg : 213
Given,
Total voltage = V = 3 * 2 = 6V
Total Resistance = R = 5+8+12
= 25 ohms
Ammeter reading = I = V/R
= 6/25 = 0.24A
Voltmeter reading = V = I * R
(which is connected = 0.24 * 12
across 12W resistor) =
2.88V

Redraw the circuit
of prev. qn, putting
in an ammeter to
measure
the current
through the
resistors and a
voltmeter to
measure the
p.d across the 12
W resistor. What
would be the
readings in the
ammeter and the
voltmeter?
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Ammeter is joined in series and the voltmeter is
connected in parallel across 12 ohms resistor.

22. Pg : 216
a) 1/R = 1/R1 + 1/ R2 = 1/1+ 1/106
= ( 1000000 + 1) / 106
= 1000001/ 1000000
= 1.000001 = 1
There fore , R = 1 Ω
(b) 1/R = 1/R1 + 1/R2 + 1/R3
= 1/1 + 1/103 + 1/106
= (10,00,000 + 1000 + 1) / 106
= 1001001 /1000000
= 1.001001 = 1
Therefore , R= 1 Ω
Judge the
equivalent
resistance when
the following are
connected in
parallel – (a) 1
ohm and 106
ohm, (b) 1 ohm
and 103 ohm,
and 106 ohm.
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24. 4 + 8 + 12 +28 = 48 ohms
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= 4 + 8 + 12 + 24 = 48 Ohms
1 / R = ½ Ohm
=> R = 2 Ohm
R = R1 + R2 + R3 +
R4
1/R = 1/R1 + 1/R2 + 1/R3 +
1/R4

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