Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
The resistance of the bulb can be calculated using the expression
\( P_1 = \frac{V_2}{R_1} \)
\(R_1 =\frac{ V_2}{P_1} \).
Substituting the values, we get
R = \(\frac{220^2}{10} = 4840 \Omega \)
The resistance of x number of electric bulb is calculate as follows :
R =\(\frac{V}{I} = \frac{220}{5} = 44 \Omega \)
The resistance of each electric bulb is 4840 \(\Omega \)
The equivalent resistance of class bulb is given by :
\(\frac{1}{R} = \frac{1}{R1} + \frac{1}{R1}+ \frac{1}{R1} +....... up to x time \)
\(\frac{1}{R} = \frac{1}{R1}× x \)
x = \(\frac{R_1}{R} = \frac{4840}{44} = 110 \)
Hence, 110 lamps can be connected in parallel.