A resistor is an electrical device that implements electrical resistance in a circuit. It is a passive two-terminal device used to adjust signal levels, reduce current flow, divide voltages, terminate transmission lines, and bias active elements.
Resistors dissipate energy while they perform their function. For example, large resistors dissipate huge amounts of power as heat, which may be used as test loads in generators, part of motor controls, or any power distribution system.
Variable resistors can be used to adjust multiple electronic devices, like lamp dimmer, volume controller, or even sensing devices for heat, light, force, chemical activity or even force. Fixed resistors are used where there is a little variation in time, temperature or operating voltage.
Alternating voltage, produced by the alternating current, is a periodic voltage, which changes its magnitude and reverses its direction with time. It is used to produce electric power and various forms of electrical energy. An alternating voltage is used as an electrical supply in every household. Appliances like televisions, electric lamps, kitchen appliances, etc., all work when the alternating voltage reaches them.
A resistor can be connected in parallel or series in an electrical circuit. For example, suppose the resistors are connected in series in an AC circuit. In that case, the voltage drops across the individual resistors must be added to find the voltage drop in the entire circuit can be calculated. Likewise, if the resistors are connected in parallel in an electric circuit, the current across individual branches must be calculated to find the current in the entire circuit.
Consider an electrical circuit having a resistor (R) and an AC voltage supply (V). As soon as they get connected, a potential difference is generated, which varies sinusoidally.
The potential difference in the circuit is given by,
v = vm sin ωt
Where vm = amplitude of the oscillating potential difference
ω = angular frequency
Using Kirchhoff’s loop rule, the current through the resistor due to the voltage applied can be calculated, as under,
∑V (t) = 0
Using the potential difference equation as mentioned above, we can write the equation as,
vm sin ωt = i R
i = vm / R sin ωt
This can be rewritten as,
i = im sin ωt
Using Ohm’s law, we get,
im = vm / R
Ohm’s law works in both AC and DC voltage conditions. Therefore, we can use it here as well. We can observe the sinusoidal nature of both the current and voltage passing through the resistor. As a result, both quantities are in phase with each other.
p = i2 R = i2m R sin2 ωt
We can express the average value of power over the complete cycle as,
p = i2 R = i2m R sin2 ωt
Here the quantities im and R are constants. Therefore, the above equation can be represented as,
p = i2m R sin2 ωt
From the trigonometric equations, we know,
sin 2ωt = ½ (1 – cos 2ωt)
Therefore, we can write sin 2ωt = ½ (1 – cos 2ωt)
Also, cos 2ωt = 0
Therefore, we can write,
sin2 ωt = ½
Thus we can also write,
p = ½ i2m R
Note: We can express AC power as DC power if we denote the current in terms of root mean square current or effective current.
Irms = i2 = 12 im2 = im2 = 0.707im
The sum over a cycle of instantaneous current is zero. Hence, the average current is also zero. However, this does not imply there is no dissipation of energy.
We know, the energy dissipated in joules is given by, i2 R. This value will always be positive as i2 will always be positive. Therefore, the thermal energy developed in the resistor from the time t to t+dt is given by,
i2 R dt = i2m (ωt) R dt
The thermal energy developed in one time period is,
U = ∫i2 R dt = ∫i2m (ωt) R dt
We need to put the limits of 0 to T, we get,
= R 0Tim2 (t) R dt
= 0Tirms2 (t) R dt
This shows that the value of the root-mean-square of the alternating current is the steady current that would generate the same amount of heat in the resistor in a given amount.
Example: The voltage, resistance and frequency of the power source in an AC circuit is 100V, 1 KΩ and 50Hz, respectively. What is the root mean square and peak value of current? How much time will the current take to reach its first negative peak?
Solution:
From the question,
Peak voltage, Vm = 100 volt
Peak current, im = Vm / R
im = 100 / 1000 A = 0.1 A
RMS value, irms = im / √2
= 0.1 / √2 = 0.070 A = 70 mA
Time taken to reach the first negative peak, t = 3T / 2 = 3 / (2×50) = 0.03 sec