#Technical studies – Study 25
In 1827 German scientist George Simon Ohm discovered the relationship between voltage, current and resistance in DC circuits. This relationship is named as ohm’s law.
Ohms law states that:-
At a constant temperature, current flowing through a conductor is directly proportional to the potential difference between the two points of the conductor and inversely proportional to the resistance of the conductor.
Where I = electric current
V = Potential difference or voltage
R = Resistance
From the above equation we can get another two equations –
i.e. V = IR
and
See Also:
Realization of ohms law using a resistor:

In the circuit above the voltage V is applied to Resistor R and the current flowing through the circuit is I. Suppose voltage V = 1 V, then according to ohms law current I = 1A. If voltage will increase then current will also increase. Therefore, the relationship between current and voltage will be a straight line as drawn below.

Power in a dc circuit:
In a dc circuit the flow of current occurs due to the voltage applied to it. Due to the current flow the circuit does some work in a unit time. The rate at which the circuit does the work is called as electric power. The symbol for power is P. The unit of electric power is joules/ sec or watt.
Therefore power P
We know that in a dc circuit V
Therefore work = VQ
= Vit (Q=it)
Now put the value of work in power equation —
P = = VI
Now in equation 1 if we put the value of V as IR then
P = VI = IRI =
Now in equation 1 if we put the value of I as then
P = VI = V
Therefore we got P = VI =
Energy in electrical circuit:
In a dc circuit the energy is the total power consumed in the circuit.
Electrical energy =
Problems
Problem 1:
Calculate the power rating for the circuit given below. Also calculate the energy it will consume in 24 hours.

Solution:-
In the above circuit resistance of the resistor R = 120Ω
Supply voltage V = 24V
Therefore the current flowing through the circuit
A
Power rating of the circuit
Energy consumption in 24 hours =
Problem 2:
Look the following details of a circuit –
- Six lamps of 25 watts working for 7 hours per day.
- One 1200 watts heater working for 4 hours per day.
- Three electric bulbs of 80 watts each working for 2.5 hours per day.
As per the above details calculate the total energy consumed by the circuit for the month of August. Also calculate the bill amount for the same if each unit of energy costs Rs. 2.
Total watts of lamps = 25 x 6 = 150 watts
Total watts of heater = 1200 watts
Total watts of electric bulb = 80 x 3 = 240 watts
Energy consumed by these appliances per day
= (150 x 7) + (1200 x 4) + (240 x 2.5) = 1050 + 4800 + 600
= 6450 = 6.45 kWh
Therefore total energy consumed by the circuit in the month of August
= 6.45 x 31 = 199.95 kWh
Bill for the month of August = 2 x 199.95 = 399.9 rs.
Please write in the comment box if you have any questions.
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