This document appears to be notes from an electronics or physics class. It includes definitions and formulas for electrical concepts like current, potential difference, resistance, heating effect of electric current, and electrical energy. Worked examples calculate equivalent resistance, current, and potential difference in circuits. Diagrams show circuit schematics. The document covers topics like Ohm's Law, resistors in series and parallel, batteries, fuses, and the operation of incandescent light bulbs.

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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.

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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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