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I used 1/4W resistors and electrolytic capacitor to combine a circuit originally,and found that the output voltage is less 2V,this is not my expectation ,so i replace them with 1/8W resistors and tantalum capacitor ,and the voltage become from 2V to 3V.

Compared with 1/4W resistors and electrolytic capacitor, do 1/8W resistor and tantalum capacitor consume less electricity? I surf the internet and find that 1/4W or 1/8W means the power that resistors can sustain,it doesn't mean how much power that the resistor consump.So i am confused now.

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    \$\begingroup\$ That theory is a bit unlikely, but what really happened is unanswerable without the details of your circuit, as the rules here require. \$\endgroup\$ Mar 11, 2018 at 4:07
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    \$\begingroup\$ we are confused also. you basically provided zero information. it is not possible to guess how you are connecting the components together. you also made no mention of what the circuit is supposed to do. ... -1 \$\endgroup\$
    – jsotola
    Mar 11, 2018 at 4:08
  • \$\begingroup\$ i replace them ... not same components ... not exactly same values \$\endgroup\$
    – jsotola
    Mar 11, 2018 at 4:10
  • \$\begingroup\$ A 1/4w resistor and a 1/8w resistor of the same resistance will make no difference to a circuit (assuming you don’t burn out the 1/8). Therefore, any chance must be from the cap or you did something else. Given what little info you gave, there’s no way to know. \$\endgroup\$
    – DoxyLover
    Mar 11, 2018 at 4:10
  • \$\begingroup\$ Your resistors may become extremely hot, and the resistance becomes somewhat higher, which changes the currents. \$\endgroup\$ Mar 11, 2018 at 13:03

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Without knowing the details of your circuit, it's hard to really answer your question or to help you understand whatever is confusing you. That said, power dissipation in these components isn't too hard to understand.

Resistors

schematic

simulate this circuit – Schematic created using CircuitLab

The power rating designation of a resistor is the amount of power the part can dissipate $$P = IV = I^2R = V^2 / R$$ indefinitely without causing lasting effects on its primary rating, it's resistance. That is to say, if you connect a 10 V DC source to a 1 kiloohm resistor, Ohm's law dictates a current flow of $$I = V/R = (10 V)/(1000 Ohm) = 10 mA$$ which means that resistor will be dissipating $$P = I^2R = (10 mA)^2(1000 Ohm) = 100 mW = 0.1 W$$. This is well below either 1/4 W or even 1/8 W. However, if you replace that first resistor for one with a resistance of 220 Ohm, the circuit would now draw $$I = V/R = (10 V)(220 Ohm) = 45.5 mA$$ and the resistor would need to dissipate $$P = I^2R = (45.5 mA)^2(220) = 0.46 W$$ Enough to burn either a 1/4 or 1/8 watt resistor! Note that you can compute the power directly in this simple example using the voltage of the source and the resistance, but I showed my work using current to hopefully more strongly outline the link between current and power dissipation here.

Capacitors

The story of power dissipation in a capacitor is more complex. There are Ohmic losses akin to the resistor above linked to the Equivalent Series Resistance (ESR) of the capacitor. There are also dielectric losses, but these are predominantly a high frequency phenomenon.

If you're working with DC signals, you can consider power dissipation of any healthy capacitor zero.

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