a) A student performed an experiment to find out the relation between
Voltage and Current in the Laboratory. He found that a linear relation
exists between these two. State the law which supports his finding.
(b) How is an ammeter connected in an electric circuit ?
( c) The power of a lamp is 100 W. Find the energy consumed by it in 1
minute.
(d) A wire of resistance 5 Ω is bent in the form of a closed circle. Find
the resistance between two points at the ends of any diameter of
the circle.
Answers
Answer:
A. The relationship between current, voltage and resistance is expressed by Ohm's Law. This states that the current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit, provided the temperature remains constant.
B. An ammeter is connected in series with the circuit to be measured. The ideal ammeter will have zero resistance so as not to disturb the circuit. We will find the shunt as part of the ammeter circuit.
C. Since there are 60 seconds in a minute, that means that the light bulb will consume (100 watts)*(60 seconds) or 6000 Joules.
D.:
The length of the wire between two points at the ends of a diameter of a circle is
half of the whole wire. The resistance of a conductor is directly proportional to its
length.
Thus, the two resistance 5/2 ohm will be in parallel.
Let R be the effective resistance.
1/R=2/5+2/5
1/R=4/5
R=5/4=1.25 ohm.
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Linear relation between current (I ) and voltage (V) is given by OHM's law I = V/R
Explanation:
relation between Voltage and Current by OHM's law
I = V/R
R = resistance
I = Current
V = Voltage
ammeter is connected in Series in an electric circuit
Voltmeter is connected in Parallel in an electric circuit
The power of a lamp is 100 W. Find the energy consumed by it in 1
minute.
1 min = 60 sec
Energy = power x time
100 x 60 = 6000 Joule = 6 kJoule
A wire of resistance 5 Ω is bent in the form of a closed circle
Each wire resistance between two points of diameter =5/2 Ω
As these will be in parallel hence net resistance
= 1/((1/(5/2) + 1/(5/2)))
= 5/4
5/4 Ω is the resistance between two points at the ends of any diameter of
the circle.
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