**#Technical studies – Study 25**

**#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:-**

**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.*

$I=\frac{V}{R}$

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 $R=\frac{V}{I}$

See Also:

**Realization of ohms law using a resistor: **

**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: **

**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 $=\frac{workdoneinacircuit}{time}$

We know that in a dc circuit V $=\frac{work}{Q}$

Therefore work = VQ

= Vit (Q=it)

Now put the value of work in power equation —

P = $\frac{Vit}{t}$ = VI $...........\left(1\right)$

Now in equation 1 if we put the value of V as IR then

P = VI = IRI = ${I}^{2}R$ $.............\left(2\right)$

Now in equation 1 if we put the value of I as $\frac{V}{R}$ then

P = VI = V$\frac{V}{R}=\frac{{V}^{2}}{R}$ $.............\left(3\right)$

Therefore we got P = VI = ${I}^{2}R=\frac{{V}^{2}}{R}$

**Energy in electrical circuit:**

**Energy in electrical circuit:**

In a dc circuit the energy is the total power consumed in the circuit.

Electrical energy = $Power\times time$

$=VIt={I}^{2}Rt=\frac{{V}^{2}}{R}t$

**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

$I=\frac{V}{R}=\frac{24}{120}=0.2$A

Power rating of the circuit $P=VI=24\times 0.2=4.8W$

Energy consumption in 24 hours = $P\times t=4.8\times 24=115.2Wh$

**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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