Ohm's law ( in terms of electrical tearms )

This law applies to electric to electric conduction through good conductors and may be stated as follows:

The ratio of potential difference (V) between any two points on a conductor to the current (I) flowing between them, is constant. provided the temperature of the conductor does not change.

In other words,  V/I = constant  or V/I = R

where R is the resistance of the conductor between the two points considered. Put in another way, it simply means that provided R is kept constant, current is directly proportional to the potential difference across the ends of a conductor. However, this linear relationship between V and I does not apply to all non-metallic conductors. For example, for silicon carbide, the relationship is given by $V=K{I}^m$ where K and m are constants and m is less than unity. It also does not apply to non-linear devices such as Zener diodes and voltage-regulator (VR) tubes.

---

---

Example: A coil of copper wire has resistance of 0 at 20°C and is connected to a 230-V supply. By how much must the voltage be increased in order to maintain the current consant if the temperature of the coil rises to 60°C? Take the temperature coefficient of resistance of copper as 0.00428from 0°c

Solution

As seen from page Temperature Coefficient of Electrical Resistance 

$\frac{R60}{R20}=\frac{1+60\times 0.00428}{1+20\times 0.00428}$
so R60=90 x 1.2568 / 1.0856 = 104.2 Ohm

Now, current at 20°C = 230/90 =23/9 A

Since the wire resistance has become 104.20 Ohm at 60°C, the new voltage required for keeping the
current constant at its previous value = 104.2 x 23/9 = 266.3 V
So, increase in voltage required=266.3 - 230 = 36.3 V

---

Example: Three resistors are connected in series across a 12-V battery. Thefirst resistor has a value of 1 Ohm, second has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of the circuit current.

Solution

Let the two unknown resistors be R2and R3and I the circuit current

$I^{2}R3 =12 and IR3=4$
So,$\mathrm{ \; R}3=\frac{3}{4}R{2}^2 Also I=\frac{4}{R2}$

Now,     I (1+R2+R3) = 12

Substituting the values of I and R3, we get

$\frac{4}{R2}\left(1+R2+\frac{3}{4}R{2}^2\right) =12 or 3R{2}^2 -8R2+4=0$


$R2=\frac{8\pm \sqrt{64-48}}{6} So R2=2\Omega or \frac{2}{3}\Omega$


$R3=\frac{3}{4}R{2}^2=\frac{3}{4}{2}^2 =3\Omega or \frac{3}{4}{\left(\frac{2}{3}\right)}^2=\frac{1}{3} \Omega$

So I = 2A or I = 6A

---

---