The resistance of the bulb can be calculated using the expression
P1 = V2 /R1
R1 = V2 /P1
Substituting the values, we get
\( R= \frac{(220)^2}{10}=4840\,\Omega\)
The resistance of x number of electric bulbs is calculated as follows:
\( R = \frac{V}{I}= \frac{220}{5}=44\,\Omega\)
\( The\, resistance \,of \,each\, electric\, bulb \,is \,4840\,\Omega\)
The equivalent resistance of x bulbs is given by
\(\frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_1} +\frac{1}{R_1}\pm ....upto\,x\,times\)
\( \frac{1}{R}= \frac{1}{R_1}\times x\)
\( x= \frac{R_1}{R}= \frac{4840}{44}=110\)
Hence, 110 lamps can be connected in parallel.
Answered by Shivani Kumari | 2 years agoExplain the following.
a. Why is the tungsten used almost exclusively for filament of electric lamps?
b. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
c. Why is the series arrangement not used for domestic circuits?
d. How does the resistance of a wire vary with its area of cross-section?
e. Why copper and aluminum wires are usually employed for electricity transmission?
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Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.