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Circuit problem help?
I don't really get this one:
Determine the power dissipated in the 7.0 resistor in the circuit shown in the drawing. (R1 = 3.0 , R2 = 7.0 and V1 = 16 V.)
I can't really copy the diagram, but it's a complete circuit with a resistor in series with resistors in parallel. So basically, the circuit first goes to resistor R1 then travels trhough a parallel circuit of two lines, one of which is a resistor of 3 ohms while the othe line contains R2 and another resistor of 1 ohm. Then it reconnects and goes back to the battery.
I can easily get the current here, but how do I find the power of R2 since the voltage would be different and the current might be different as well?
3 Answers
- ToneyLv 51 decade ago
OK, IF I understand your circuit, since there are only 2 lines in parallel, I assume that R2 is in series with the (1) one ohm resistor, and they are in parallel with a 3 (three) ohm resistor.
If that is correct, here are the measurements:
current thru R2 = 1.4545 Amps (16V / 11 ohms)
volts across R2 = 10.1818 Volts (7 ohms x 1.45454545 amps)
power dissipated by R2 = 14.8094 watts (10.1818V x 1.45454545 amps)
Source(s): I am an electronic engineer - 1 decade ago
you can use thevenin's theorem or norton theorem to solve this.
i am not getting the circuit you described.
but since there is only one voltage source you can go by the thevenin's theorem.
