POWER IN DC CIRCUITS

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Resistors dissipate energy in the form of heat, and the rate at which they dissipate energy is called power. The power dissipated by resistors is delivered by the voltage and/or current sources in the circuit.

The unit of power is the watt (one joule/second).

There are several ways of calculating the power of resistors.

Example 1

Find the power of each circuit element if V=150 V and R = 1 kohm.

First, find the current of the network:

I = V /(R+2*R) =150/(1+2) = 150/3 = 50 mA

The power of the resistors is then:

P1 = I2 * R = 502 *10-6 *103 = 2.5 W;

P2 = I2 * 2*R = 502 *10-6 * 2*103 = 5 W;

 

The power delivered by the voltage source is:

PV = – I * V = – 5 * 10-2 *150 = -7.5 W.

Note that the current is opposite to the voltage in the source. By convention in this case, power is denoted as a negative quantity. If a circuit contains more than one source, some sources may actually dissipate energy if their current and voltage have the same direction.

The solution using TINA’s DC Analysis:


Click/tap the circuit above to analyze on-line or click this link to Save under Windows

The simulation results agree with the calculated powers:

{Solution by TINA’s Interpreter!}
I:=V/(R+2*R);
P1:=I*I*R;
P2:=2*R*I*I;
P1=[2.5]
P2=[5]
PV:=-I*V;
PV=[-7.5]
#Solution by Python
I=V/(R+2*R)
P1=I*I*R
P2=2*R*I*I
print(“P1= %.3f”%P1)
print(“P2= %.3f”%P2)
PV=-I*V
print(“PV= %.3f”%PV)

 

We can calculate the power dissipated by each resistor if we know either the voltage or the current associated with each resistor. In a series circuit, it is simpler to find the common current, while in a parallel circuit it is easier to solve for the total current or the common voltage.

Example 2

Find the power dissipated in each resistor if the source current is I = 10 A.

In this example, we have a parallel circuit. To find the power we must calculate the voltage of the parallel circuit:

 

Find the power in each resistor:

 

 

Solution using TINA’s DC Analysis


Click/tap the circuit above to analyze on-line or click this link to Save under Windows

The simulation results agree with the calculated powers.

 

Solution by TINA’s Interpreter
V:=I*Replus(R1,R2);
V=[120]
I1:=I*R2/(R1+R2);
I1=[4]
I2:=I*R1/(R1+R2);
I2=[6]
P1:=R1*sqr(I1);
P1=[480]
P2:=R2*sqr(I2);
P2=[720]
Ps:=-V*I;
Ps=[-1.2k]

Example 3

Find the power in the 5 ohm resistor.


Click/tap the circuit above to analyze on-line or click this link to Save under Windows

Solution using the Interpreter in TINA
I:=Vs/(R1+Replus(R2,R2));
I=[1]
P5:=I*I*R1;
P5=[5]
#Solution by Python
Replus= lambda R1, R2 : R1*R2/(R1+R2)
I=Vs/(R1+Replus(R2,R2))
V1=Vs*R1/(R1+Replus(R2,R2))
P1=I*V1
print(“P1= %.3f”%P1)

 Example 4


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Find the power in the resistor RI.

Solution using TINA’s Interpreter
Ir:=I*R/(R+R1+replus(R2,(R3+RI)))*R2/(R2+R3+RI);
Ir=[1.25m]
PRI:=sqr(Ir)*RI;
PRI=[125m]
#Solution by Python
Replus= lambda R1, R2 : R1*R2/(R1+R2)
Ii=I*R/(R+R1+Replus(R2,(R3+RI)))*R2/(R2+R3+RI)
Pi=Ii**2*RI
print(“Pi= %.3f”%Pi)


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