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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:

P_{1} = I^{2} * R = 50^{2 }*10^{-6 }*10^{3} = 2.5 W;

P_{2 }= I^{2} * 2*R = 50^{2 }*10^{-6 }* 2*10^{3} = 5 W;

The power delivered by the voltage source is:

P_{V} = – 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:

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]

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

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.

Solution using the Interpreter in TINA

I:=Vs/(R1+Replus(R2,R2));

I=[1]

P5:=I*I*R1;

P5=[5]

**Example 4**

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]