A: Total power dissipated by three identical bulb of rating (60 W, 220 V) in given
network is 180 W.
220 V
R: Heat dissipated by resistance is directly proportional to square of current
passes through it.
Answers
Answer:
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Explanation:
The circuit can be reduced, step by step, to a single equivalent resistance. The 8 ohm and the 8 ohm are connected in parallel , and so they can be replaced by an equivalent resistor Rp of 4 ohms, using the below equation.
Rp=
(8+8)
(8×8)
=4ohms
This resistor is connected in series with the 4 ohm resistor R1. The total resistance Rt of the circuit is then
Rt =R1+Rp=4+4 = 8 ohms.
Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get,
I=
R
V
=
8
8
=1A.
Hence, the current flowing through the resistor R1 (4 ohms) is 1A.
Now the power dissipated in the resistor R1 (4 ohms) is given as follows.
P=I
2
R=1
2
×4=1×4=4W.
Hence, the power dissipated in the resistor R1 (4 ohms) = 4W.