Hello everyone, I will discuss to you how to get the power dissipated by each resistor in a series circuit.
When resistors are connected in series, the current through each resistor is the same. In other words,
the current is the same at all points in a series circuit.
When resistors are connected in series, the total voltage (or potential difference) across all the resistors
is equal to the sum of the voltages across each resistor.
The easiest way to calculate power dissipated by a resistor in a series circuit is to use Joule’s law, P = IV,
where P is electric power. In this case, each resistor has the same full current flowing through it. By
substituting Ohm’s law V = IR into Joule’s law, we will get the power dissipated by each resistor in a
series circuit.
So the formula that we will use is,
P 1 = I² R1, P2 = I² R2, …. Pn= I² Rn (PDF)
The n here is denoting an unspecified member of a series of numbers. (It means pwede siya maging,
1,2,3, so on and so forth. depende kung pila ka resistor ang ara sa given)
Remember, P represents power, I is for current and R is for Resistance.
(Diba sa formula ta, we need the current to get the power dissipated by each resistor in a series circuit.)
Here are the steps on how to get the power dissipated by each resistor in series.
1.Find the total resistance
(para ma kwa naton ang current, need ta danay i-solve ang total resistance)
(The total resistance is simply the sum of the individual resistances, using this formula)
Rt= R1+ R2 +…. Rn (I will show you later sa example kung paano ni sa i-solve and I think nadumdoman
niyo man ni kay gin discuss na ni sang previos group)
2. Find the current.
The current is found using Ohm’s law, V = IR. Entering the value of the applied voltage and the total
resistance yields the current for the circuit:
The formula that we will use is:
(current is equal to voltage over total resistance)
I=
𝒗
𝑹𝒕
3. Calculate the power dissipated by each resistor.
We can calculate the power dissipated by each resistor by using the formula:
(power is equal to current squared times the individual resistances
P 1 = I² R1, P2 = I² R2, …. Pn= I² Rn
Example 1:
R1
R2
R3
Here is example 1, Suppose the voltage output of the battery in Figure 1 is 12 V, and the resistances are
R1 = 2Ω,R2= 8 Ω, and R3 = 16Ω
Given:
V= 12 V
R1= 2 Ω
R2= 8Ω
R3= 16 Ω
Rt= ?
I= ?
P1= ?
P2= ?
P3= ?
Remember, the first step is to find the total resistance, so using the formula
Rt= R1+ R2 +…. Rn (just substitute the r1, r2, and r3 base sa given.
Rt= 2 Ω + 8 Ω + 16 Ω
Rt= 26.0 Ω
Second is to find the current.
V represent voltage and sa given ta It is 12 volts and the total resistance is 26 ohms
I=
I=
𝒗
𝑹𝒕
𝟏𝟐𝒗
𝟐𝟔𝜴
I =0.4615 A
(And to get the power dissipated by each resistor,
Use the formula, power is equal to current squared times the individual resistances)
I represents current
Ang value sang current ta is 0.4615 so i substitute ta lang
And kay gina kwa ta ang power dissipated sa resistor 1, and value man sang resistor 1 ang ibutang ta di
P1 = I2 R1 = (0.4615 A)2(2 Ω) = 0.4260 W
The power dissipated in resistor 1 is 0.4260 W
Same man diri, although lain lang ang value sang resistor
P2 = I2 R2 = (0.4615 A)2(8 Ω) = 1.7039 W
The power dissipated in resistor 2 is 1.7039 W
P3 = I2 R3 = (0.4615 A)2(16 Ω) = 3.4077 W
The power dissipated in resistor 3 is 3.4077 W
Practice Problem:
R1
R2
R3
Given:
V= 125 V
R1= 10 Ω
R2= 15Ω
R3= 25 Ω
Rt= ?
I= ?
P1= ?
P2= ?
P3=?
Rt= R1+ R2 +…. Rn
Rt= 10 Ω + 15 Ω + 25 Ω
Rt= 50 Ω
I=
𝒗
𝑹𝒕
I=
𝟏𝟐𝟓𝒗
𝟓𝟎𝜴
I = 2.5 A
P 1 = I² R1, P2 = I² R2, …. Pn= I² Rn
P1 = I2 R1 = (2.5 A)2(10 Ω) = 62.5W
P2 = I2 R2 = (2.5 A)2(15 Ω) =93.75 W
P3 = I2 R3 = (2.5 A)2(25 Ω) =156.25 W
That is how you get the power dissipated in series.